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46 Objective MHT-CET Chemistry

A true solution : • A true solution is formed when soluble substances are dissolved in the solvent. The sizes of the particles dissolved are very small of the order of 10–8 cm. True solutions are homogeneous and cannot be separated into components by simple mechanical methods.

A true solution is defined as a homogeneous f

mixture of two or more substances, the composition of which is not fixed and may be varied within certain limits.A solution is formed by two components, solvent f

and one or more solutes. The component of the solution which constitutes larger part of the solution is called solvent and the other component that constitutes smaller part is called solute.

Homogeneous solution : • Solution is homogeneous if its composition is uniform throughout the body of the solution.Heterogeneous solution : • Solution is heterogeneous when two or more phases are present in it.Solvation : • The process of interaction of solvent molecules with solute particles to form aggregates. When water is a solvent, the process of solvation is called hydration or aquation.

The extent to which solute dissolves in solvent to f

form homogeneous solution depends on nature of solute and solvent.The general rule is, like dissolves like f i.e., polar

2 Solutions and Colligative Properties

2.1 Introduction

2.2 Types of Solutions

2.3 Concentration of Solutions of Solids in Liquids

2.4 Solubility of Gases in Liquids

2.5 Solid Solutions

2.6 Colligative Properties

2.7 Lowering of Vapour Pressure

2.8 Boiling Point Elevation

2.9 Freezing Point Depression

2.10 Osmosis and Osmotic Pressure

2.11 Abnormal Molecular Masses

2.12 van't Hoff Factor

2.1 IntroductionTwo-third of the earth is covered with water, the most essential, vital component for life. Water in the sea is not in pure form. It contains sodium chloride, magnesium chloride, calcium salts and gases like oxygen, carbon dioxide in dissolved state. Thus, sea water is a ‘solution’. Solution consists of at least two components, a medium and dissolved phase. The physical properties of solution are mainly originated by intermolecular forces of attraction between the solvent molecules. These intermolecular forces of attraction are changed by the presence of dissolved substances. The physical properties include vapour pressure, boiling point, freezing point and osmotic pressure.

SolutionsSolutions are mixtures of two or more components. Depending on sizes of the components, the mixtures are classified into three types :

A coarse mixture : • It is formed when the sizes of the constituent components are relatively bigger, e.g.; mixture of salt and sugar.A colloidal dispersion : • It is formed when the sizes of the particles dispersed in solvent are in the range of 10–7 cm to 10–4 cm. Colloidal particles carry positive or negative charge which stabilizes colloidal dispersion e.g.; ferric hydroxide sol, arsenic sulphide sol, etc. Colloidal solutions are heterogeneous and can be easily separated.

47Solutions and Colligative Properties

solutes are soluble in polar solvents, (e.g.; NaCl in water) while non-polar solutes are soluble in non-polar solvents (e.g.; iodine in CCl4).

Water is called • universal solvent, as it is polar and has very high dielectric constant, hence, dissolves most of the polar solutes.Solutions containing two, three or four components •are called binary, ternary and quaternary solutions respectively.Solutions prepared in water are called • aqueous solutions and solutions in other solvents are called non-aqueous solutions.As the solute particles are very small, the components •of true solutions cannot be separated by simple physical methods like centrifugation, filtration, etc.


Solvent Solute ExamplesSolid Solid Alloys like brass, bronze, copper in

gold etc.Solid Liquid Amalgams of mercury with metals

Solid Gas Hydrogen gas in palladium metal, pumice stone

Liquid Solid Iodine in CCl4, benzoic acid in C6H6, sugar in water

Liquid Liquid Ethanol in waterLiquid Gas Oxygen, carbon dioxide in water

Gas Solid Iodine in airGas Liquid Chloroform in nitrogenGas Gas Air, mixtures of non-reacting gases


The concentration of a solution is defined as the amount •of solute dissolved in a specific amount of solvent.Solutions containing relatively less amount of solute are •called dilute solutions and if it contains relatively more amount of solute then the solution is called concentrated solution.Concentration of solutions may be expressed in different •ways as discussed below :

Name Symbol Formula Definition Effect of temperature

Percentage by mass

%(w/W) Mass of soluteTotal mass of solution

100×Amount of solute in grams present in 100 g of solution.

No effect

Volume percentage

%(v/V) Volume of soluteTotal volume of solution

×100Volume of solute in mL dissolved in 100 mL of the solution.

Changes with change of temperature.

Mass by volume percentage

% (w/V) Mass of soluteTotal volume of solution in mL

100×Amount of solute in grams dissolved in 100 mL of the solution.


Strength g/L (or g/dm3)

Mass of solute in gramsVolume of solution in L (or dm )3

Amount of solute in grams present in one litre (or dm3) of solution.


Parts per million

ppm Mass or volume of soluteTotal mass or volume of solution

×1006 Number of parts of solute present in million (106) parts of solution.

No effect

Molarity M Moles of soluteVolume of solution in L(or dm3 )

Number of moles of solute dissolved in one litre (or one dm3) of solution.


Molality m Moles of soluteMass of solvent in kg

Number of moles of solute dissolved in 1 kg of the solvent.

No effect

Mole fraction

xx

n nAA

A B=

+n Ratio of number of moles of

one component to the total number of moles of all the components.

No effect


Illustration : 1.23 g of sodium hydroxide (molar mass = 40) are dissolved in water and the solution is made to 100 cm3. Calculate the molarity of the solution.Soln.: Amount of NaOH = 1.23 gVolume of solution = 100 cm3

Moles of NaOH = Mass of NaOHMolar mass of NaOH

= 1 2340. = 0.0307

Molarity = ×

= × =

Moles of NaOHVolume of solution

M

1000

0 0307100

1000 0 31. .

Illustration : 1.8 g of glucose (molar mass = 180) are dissolved in 60 g of water. Calculate (a) the molality (b) mole fraction of glucose and water.Soln.: (a) Molality of solution

= Moles of soluteMass of solvent in g

× 1000

\ Molality = Moles of glucoseMass of water

× 1000

= 1 8 1000180 60. ×

× = 0.167 m

(b) Mole fraction

= Moles of soluteMoles of solute + Moles of solvent

Moles of glucose = =1 8180

0 01. . ;

Moles of water = =6018

3 33.

Mole fraction of glucose =+

=0 010 01 3 33

0 003.. .

. ;

Mole fraction of water =+

=3 330 01 3 33

0 997.. .

.

Illustration : 4 g of sodium chloride was dissolved in 300 g of water. Calculate percentage by mass of sodium chloride in solution.Soln.: Percentage by mass of sodium chloride (w/W)

= mass of sodium chloridemass of sodium chloride + mass of wwater

×100

=+

× = =4

4 300100 400

3041 316

gg g

by mass. %

Depending upon the quantity of solute dissolved in a •liquid solvent, solutions can be of three types :

Saturated solution : f A solution which cannot dissolve any further amount of solute at a given temperature is called saturated solution.

Unsaturated solution : f A solution in which more amount of the solute can be dissolved at a given temperature is called unsaturated solution.Supersaturated solution : f A solution in which amount of solute is more than it can dissolve at a particular temperature is called supersaturated solution.

Solubility • of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature.Factors affecting solubility of a solid in a liquid : •

Effect of temperature : f If the dissolution process is endothermic −(DsolH > 0), the solubility increases with rise in temperature.If dissolution process is exothermic −(DsolH < 0) the solubility decreases with rise in temperature.

Effect of pressure : f Pressure does not have any significant effect on solubility of solids in liquids as these are highly incompressible.

2.4 Solubility of Gases in LiquidsGases are soluble in liquids including water. The solubility of gases like O2, N2, etc. are much low. O2 molecules being non-polar, have less solubility in polar solvent, water. CO2 and NH3 gases are more soluble in water as CO2 reacts with water to form carbonic acid and NH3 reacts with water to form ammonium hydroxide. HCl gas is polar, its solubility is very high in water, forming hydrochloric acid.

Factors affecting solubility of a gas in a liquid : •Effect of pressure : f The solubility of gas increases with the increase in external pressure.Henry’s law - It states that the solubility of a gas in a liquid at constant temperature is proportional to the pressure of the gas above the solution.

S ∝ P i.e., S = K⋅Pwhere S is the solubility of the gas in mol dm–3, P is the pressure of the gas in atm, K is constant of proportionality and has the unit of mol dm–3 atm–1. If P = 1 atm, then S = K. Hence, Henry’s constant K is defined as solubility of gas in mol dm–3 at 1 atmospheric pressure at reference temperature. If several gases are present then the solubility of any gas may be evaluated by using P as partial pressure of that gas in the mixture.Effect of temperature : f According to Charles’ law, volume of a given mass of a gas increases with increase of temperature. Therefore, volume


of a given mass of dissolved gas in solution also increases with increase of temperature, so that it becomes impossible for the solvent in solution to accommodate gaseous solute in it and gas bubbles out. Hence, solubility of gas in liquid decreases with increase of temperature.Effect of addition of soluble salt : f Solubility of dissolved gas is suppressed when a soluble salt is added to the solution of gas.

Illustration : The solubility of nitrogen gas at 1 atm pressure at 25°C is 6.8 × 10–4 mol dm–3. Calculate the solubility of N2 gas from atmosphere at 25°C if atmospheric pressure is 1 atmosphere and partial pressure of N2 gas at this temperature and pressure is 0.78 atm.Soln.: (i) S = 6.8 × 10–4 mol dm–3, PN2

= 1 atm S = K⋅PN2

6.8 × 10–4 mol dm–3 = K × 1 atm K = 6.8 × 10–4 mol dm–3 atm–1

(ii) P = 0.78 atm, S = ?S = 6.8 × 10–4 mol dm–3 atm–1 × 0.78 atm = 5.304 × 10–4 mol dm–3

i.e. Solubility of N2 is reduced to = 5.304 × 10–4 mol dm–3

2.5 Solid SolutionsA solid solution of two or more metals or of a metal or •metals with one or more non-metals is called an alloy or solid solution.All the properties of the pure metals are improved when •they form solid solutions, i.e., alloys.

Duralumin (Al + Cu + Mg + Mn) :Light and strong as steel. −Used in the construction of aircrafts. −

Aluminium bronze (Al + Cu + Mn)Lead alloy (Pb + 10 - 20% Sb) :

Acid resistant. −Used for bearings, bullets, shrapnel and for −manufacturing lead storage battery plates.

Babbitt metal (Sb + Sn + Cu) :Antifriction alloy. −Used in machine bearings. −

Stainless steel (Steel + Cr + Ni) :Resistant to corrosion. −Used in cutlery. −

Spiegeleisen (5-20% Mn in Iron ) and Ferromanganeous (70 - 80% Mn + 30 - 20% Fe) :

Used for making very hard steels and to manufacture −rails, safes and heavy machinery.

Manganin (84% Cu + 12% Mn + 4% Ni) :Has almost zero temperature coefficient of −electrical resistance.Used for making electrical measurements. −

Amalgams (Hg + metals) :Used for extracting metals from ores. −

2.6 Colligative PropertiesThese are the properties that depend on the number of solute particles in solution and not on the nature of the solute particles. These are :

Lowering of vapour pressure, •Elevation of boiling point of solvent in solution, •Depression of freezing point of solvent in solution and •Osmotic pressure. •

2.7 Lowering of Vapour PressureVapour pressure of liquids : •

The f vapour pressure of a substance is defined as the pressure exerted by the gaseous state of that substance when it is in equilibrium with the solid or liquid phase.Vapour pressure of a liquid, increases with the f

increase of temperature.The f boiling point of a liquid is a temperature at which vapour pressure of liquid becomes equal to external pressure. If the boiling is carried out in an open atmosphere then external pressure is the atmospheric pressure.

Vapour pressure lowering : •The vapour pressure of a liquid solvent is lowered f

when a non-volatile solute is dissolved in it to form a solution.This is due to the fact that in case of pure solvent, f

its surface area is completely occupied by volatile solvent molecules. While in case of solution of non- volatile solute, its surface area is not completely available for volatile solvent; partly it is occupied by non-volatile solute.Hence, rate of evaporation of the solution will be f

less as compared to that of pure solvent and vapour pressure of solution is lower than that of the pure solvent.If f p1° is the vapour pressure of pure solvent and p is the vapour pressure of the solution of non-volatile


solute in the same solvent, then p < p1° and the lowering of vapour pressure is, Dp = p1° – pThe difference between vapour pressure of pure f

solvent and the vapour pressure of solvent from solution is called vapour pressure lowering.The ratio of vapour pressure lowering of solvent f

from solution to the vapour pressure of pure solvent is called the relative lowering of vapour pressure.

Dpp

p pp1

1

1°=

° −°

Raoult’s lawThe law states that, the partial vapour pressure of any •volatile component of a solution is the product of vapour pressure of that pure component and the mole fraction of the component in the solution.

p1 = p°1x1 and p2 = p°2x2 where p°1 and p°2 are vapour pressures of pure components 1 and 2 respectively, at the same temperature.Total vapour pressure, • ptotal of solutions of two volatile components is the sum of partial vapour pressures of the two components,

ptotal = p1 + p2 = x1 p1° + x2 p2° = (x1)p1° + (1 – x1) p2° = p2° + (p1° – p2°)x1 The solution which obeys Raoult’s law over the entire

range of concentration is called an ideal solution. If a solution does not obey Raoult’s law, the solution is non-ideal.If • y1 and y2 are the mole fractions of the components 1 and 2 respectively in the vapour phase then,

p1 = y1 ptotal and p2 = y2 ptotal

Raoult’s law for a solution of non-volatile solute

Non-volatile solute does not evaporate and does not •contribute to the total vapour pressure of solution. The vapour pressure of pure component A1 is p1° and that of component A2 is p2° = 0.

p = p2° + (p1° – p2°)x1 (as p2° = 0) p = 0 + (p1° – 0)x1 i.e., p = p1°x1 …(1)

The lowering of vapour pressure • Dp is given by, Dp = p1° – p = p1° – p1°x1 = p1°(1 – x1) But 1 – x1 = x2 Hence, Dp = p1°x2 …(2)

Lowering of vapour pressure is the product of vapour •pressure of pure solvent and mole fraction of non-volatile solute dissolved in volatile solvent to form a

solution. Thus the lowering of vapour pressure depends on nature of pure solvent and concentration of solute in mole fraction.

Now the relative lowering of vapour pressure is given by,

Dp

p

p p

p

p x

px

1

1

1

1 2

12°

°

°

°

°=

−= =

Hence, relative lowering of vapour pressure = x2 Equation (2) proves that the lowering of vapour

pressure is a colligative property because it depends on the concentration of non-volatile solute.

Molar mass of solute and relative lowering of vapour pressure

Dp

p

p p

px

nn n

W MW M W M1

1

12

2

1 2

2 2

1 1 2 2°

°

°=

−= =

+=

+/

/ / •

For dilute solutions n1 >> n2

\ Dp

p

nn

W MW M

W MW M1

2

1

2 2

1 1

2 1

1 2°= = =

//

Knowing the masses of non-volatile solute and the solvent in dilute solutions and by determining experimentally vapour pressure of pure solvent and the solution it is possible to determine molar mass of a non-volatile solute.

Illustration : Calculate the weight of a non-volatile solute (mol. wt. 40), which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.Soln.: Given: p = (80/100)p°, M2 = 40W1 = 114 g, M1 = 114

We know, p pp

WM

MW

°−°

= ×2

2

1

1

\ p pp

W°− °°

= ×( / )80 100

40114114

2

\ W2 = 18 g (weight of solute required)

2.8 Boiling Point ElevationBoiling point • is defined as the temperature at which the vapour pressure of liquid becomes equal to the atmospheric pressure.

It is a characteristic property of liquids and is a f

criterion to check the purity of liquid.It increases with increase in external pressure. f

Liquids having greater intermolecular forces have f

high boiling points.


Depression in freezing point and molar mass of the •solute :

DTf =××

KWM Wf

B

B A

1000 or MK W

T WBf B

f A=

× ××

1000

D

Illustration : The freezing point of a solution containing 50 cm3 of ethylene glycol in 50 g water is found to be –34°C. Assuming ideal behaviour, calculate the density of ethylene glycol.(Kf for H2O = 1.86 K kg mol–1)Soln.: Given: Vethylene glycol = 50 cm3, W1 = 50 gKf(H2O)

= 1.86 K kg mol–1, M2 = 62 (glycol)DTf = 34°C, W2 = 50 × d (glycol)

We know, DTf = 1000 2

2 1

× ×

×

K W

M Wf

Substituting values, 34 = 1000 1 86 5062 50

× × ××

. ( )d

\ d = 1.133 g/cm3

2.10 Osmosis and Osmotic PressureOsmosis : • The net spontaneous flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane.Osmotic pressure : • The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane.

Two or more solutions having same osmotic f

pressure at a given temperature are called isotonic solutions.If one solution is of lower osmotic pressure, f

it is called hypotonic with respect to the more concentrated solution. The more concentrated solution is said to be hypertonic with respect to the dilute solution.If a pressure higher than the osmotic pressure f

is applied on the solution, the solvent will flow from the solution into the pure solvent through the semipermeable membrane and the process is called reverse osmosis. It is used in desalination of sea water.

Laws of osmotic pressurevan’t Hoff–Boyle’s law : • It states that, at constant temperature the osmotic pressure (p) of a dilute solution is directly proportional to its molar concentration or inversely proportional to the volume of the solution.

At constant T, p ∝ C (in mol L–1)

Elevation of boiling point : • Solution has lower vapour pressure and hence f

higher boiling point than pure solvent. The increase in the boiling point, DTb = Tb – Tb° is known as elevation of boiling point.For dilute solutions, f DTb ∝ m

or DTb = Kb m = KWM Wb

B

B A

××

1000

or MW K

T WBB b

b A=

× ××

1000D

where m is molality of solution and Kb is called boiling point elevation constant or molal elevation constant or ebullioscopic constant, having unit K kg mol–1.

Illustration : The molal elevation constant for water is 0.51 K kg mol–1. Calculate the b. pt. of solution made by dissolving 6 g urea in 200 g water.Soln.: Given: W2 = 6 g, M2 = 60 (urea), W1 = 200 g,Kb = 0.51 K kg mol–1

We know,

DTb = 1000 2

2 1

× ××K W

M Wb

Substituting values, DTb = 1000 0 51 660 200× ××

. = 0.255°C

As elevation in b. pt. = 0.255°C\ B. pt. of solution = 100 + 0.255 = 100.255°C


Freezing point • of a liquid is the temperature at which the vapour pressure of solid is equal to the vapour pressure of liquid.Depression of freezing point : • The lowering of vapour pressure of a solution causes a lowering of the freezing point compared to that of the pure solvent. The decrease in freezing point, DTf = Tf° – Tf is known as depression in freezing point.Freezing point depression and vapour pressure •lowering : For dilute solutions,

DTf ∝ p1° – p DTf ∝ m or DTf = Kf m where Kf is known as freezing point depression constant

or molal depression constant or cryoscopic constant, having unit K kg mol–1.


If n moles of solute is dissolved in V litres then,

C nV

=

\ p p∝ ∝nV V

or 1 for constant n

\ pV = constant or pC

= constant

van’t Hoff-Charles’ law : • It states that, the concentration remaining constant, the osmotic pressure of a dilute solution is directly proportional to the absolute temperature.

At constant C, p ∝ T i.e., pT

= constant

van’t Hoff general solution equation : • p = CRT R = 8.314 J K–1 mol–1 if p is in N m–2 and V in m3

R = 0.082 L atm K–1 mol–1 if p is in atm and V in dm3

Determination of molar mass from osmotic •pressure :

p = nV

RT2 , p = WM

RTV

2

2, M2 = W RT

V2p

van’t Hoff–Avogadro’s law : • It states that, two solutions of equal concentrations of different solutes exert same osmotic pressure at the same temperature. It can also be, stated as, equal volumes of isotonic solutions contain an equal number of solute particles at the given temperature.

For a given solution pV = nRT p1V1 = n1RT1 for solution 1 …(i) p2V2 = n2RT2 for solution 2 …(ii) If p1 = p2, T1 = T2 and V1 = V2 then from equations (i)

and (ii), n1 = n2 Since, number of moles are equal, number of molecules

are also equal. Hence, osmotic pressure and temperature remaining

the same, equal volumes of solutions would contain equal number of moles of the solute.

For isotonic solutions of equal volume, n1 = n2


For substances undergoing association or dissociation •in solution, the molecular mass determined by studying any of the colligative properties is different than the theoretically expected value, and the substance is said to show abnormal molecular mass.Dissociation of electrolyte solutes : • Electrolytic solutes when dissolved in solvent dissociate to produce

multiple number of ions/particles. This results in increase of number of solute particles in solution which in turn results in increase of colligative properties and increase of DTf /m value and the value of DTf /m is approximately equal to integral multiple of Kf value. The value of integral is equal to total number of ions produced on dissociation as shown :

(i) HCl → H+ + Cl–; 2 particles; DTf /m = 2 × 1.86 K mol–1 kg (ii) NH4Cl → NH4

+ + Cl–; 2 particles; DTf /m = 2 × 1.86 K mol–1 kg (iii) CoCl2 → Co2+ + 2Cl–; 3 particles; DTf /m = 3 × 1.86 K mol–1 kg (iv) K2SO4 → 2K+ + SO4

2–; 3 particles; DTf /m = 3 × 1.86 K mol–1 kg (v) AlCl3 → Al3+ + 3Cl–; 4 particles; DTf /m = 4 × 1.86 K mol–1 kg DTf /m = Kf value observed in case of solutions of

electrolytes may not be exactly two fold, three fold etc of theoretical Kf value observed in case of solutions of non-electrolyte solute. The value fluctuates with degree of dissociation of solute in solution.Association of solutes : • In some non-polar solvents, two or more molecules of solute associate to form bigger molecules. For example, in benzene solutes like acetic acid, benzoic acid, etc. associate to form dimers. This association is due to the hydrogen bonding between these molecules.

2CH3COOH = (CH3COOH)2

2C6H5COOH = (C6H5COOH)2

Hence, numbers of solute particles are reduced to almost half. Observed molecular masses of these species are almost twice the expected values in dilute solutions. Due to association the total number of molecules in solution will be almost half of the number of molecules of the substance dissolved. Hence such solutions show abnormally low colligative properties. The observed molecular masses are almost double.

2.12 van’t Hoff Factor

The colligative properties of electrolytes may be •expressed in relation to colligative properties of non-electrolytes by using van’t Hoff factor i. It is defined as the ratio of observed colligative property produced by a given concentration of electrolyte solution to the property observed for the same concentration of non-electrolyte solution.


i =Observed value of the colligative property

Theoretical vallue of the colligative property

or iMM

th

o= =

Theoretical molecular massObserved molecular mass

i =

Total number of moles of particles after association dis/ ssociation

Number of moles of particles before association / ddissociation

,

for association, i < 1; for dissociation, i > 1van’t Hoff factor ( • i) and degree of dissociation (a) :

a = −−

in

11

;

i

MM

n= = + −(theoretical)(observed)

[ ( ) ]1 1 a

a =

−−

M MM n

(theoretical) (observed)(observed) ( )1

van’t Hoff factor ( • i) and degree of association (a) :

a = −−1

1 1in/

Modified equations for colligative properties : •

p pp

inn

1 1

1

2

1

°−°

= ⋅ , DTb = iKb m, DTf = iKf m, p = i n2RT/V

Illustration : A decimolar solution of potassium ferrocyanide is 50% dissociated at 300 K. Calculate the osmotic pressure of the solution (R = 8.314 JK–1 mol–1).Soln.: Initial number of moles of K4[Fe(CN)6] = 0.1, (as the solution is decimolar).a = degree of dissociation = 50% = 1/2Thus, we may represent the facts as below : K4[Fe(CN)6] → 4K+ + [Fe(CN)6]4– totalInitial no. 0.1 0 0 0.1of moles After 0.1(1 – a) 0.1 (4a) 0.1 (a) 0.1(1 + 4a)dissociation = 0.1 [1 + (4 × 1/2)] = 0.3

van’t Hoff factor,

i = Number of particles after dissociationActual number of partticles before dissociation

= 0.30.1

= 3pV = inRT

p = inRTV

= iCRT (C = n/V = molarity = 0.1 mol L–1)

or 0 1

10 3 3. mol

m− = 100 mol m–3

or, p = 3 × 100 × 8.314 × 300 = 74.826 × 104

= 7.4826 × 105 Nm–2 = 7.4826 atm


van

’tH

off

facto

r:

Degre

eofD

isso

cia

tion

an

dA

ssocia

tion

:

Mo

dif

ied

Eq

ua

tio

ns

of

Co

llig

ati

ve

Pro

perti

es

:

TiK

mb

b=

TiK

mf

f=

=

/in

RTV

2

SO

LU

TIO

NS

AN

D C

OL

LIG

AT

IVE

PR

OP

ER

TIE

S

within

Moles

Moles

Isoto

nic

Solu

tion

s:Solutionshaving

sameosm

oticpressure.

Hy

po

ton

icS

olu

tio

ns

:Solutions

having

lower

osmoticpressure

than

theother.

Hyp

erto

nic

Solu

tion

s:Solutions

having

higherosmoticpressurethan

theother.

Observedvalueofcolligativeproperty

Theoreticalvalueofcolligativeproperty

i

Theoreticalvalueo

fmolecularmass

Observedvalueo

fmolecularmass

(dissociation)

1 1

i n

(association)

1 11/

i n

11

2

11

p

pin

pn

pp

xA

AA

=°

pp

xB

BB

=°

Law

sofO

smoti

cP

ress

ure

:van

’tH

off

-Boyle

’sla

w:

Atconstant

,=constant

TV

or=constant

van

’tH

off

-Ch

arle

s’la

w:

Atconstant

,=constant

C

van

’tH

off

gen

erals

olu

tion

equ

ati

on

:

=

CRT

van

’tH

off

-Avogad

ro’s

law

:Forisotonicsolutionsofequalvolum

e,n

n1

2=

.

1(

)in

mol

L

C

T

x

x

KK

KK

Hen

ry’s

Law

:It

statesthat

the

solubility

ofagasin

aliquid

atconstanttemperatureisproportionalto

thepressure

ofthegasabovethe

solution.

SP

i.e.

SK

P,

=

Massofsolute

100

Totalm

asso

fsolution

Mass

Perc

en

tage

:(/)%

ofsolution=

wW M

assofsolute

100

Totalvolum

eofsolutioninmL

Mass

by

volu

me

perc

en

tage

:(

/)%

ofsolution=

WV

Volum

eofsolute

100

Totalvolum

eofsolution

Volu

me

Perc

en

tage

:(/)%

ofsolution=

vV

Osm

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2.1 Introduction

1. The mixture of salt and sugar is called a(a) coarse mixture(b) homogeneous mixture(c) racemic mixture(d) solution mixture.

2. In a solution, the larger proportion of the component is known as(a) solution (b) solute(c) solvent (d) mixed solution.

3. A solution is defined as a(a) homogeneous mixture of two or more substances(b) heterogeneous mixture of two or more substances(c) homogeneous mixture of liquid and solid

components only(d) homogeneous mixture consisting water as one

of the components.

4. The substances involved in solution are called(a) mole (b) atoms(c) molecules (d) components.

5. The scientist who won the first Noble prize in chemistry in the year 1901, for his work in solutions was(a) van’t Hoff (b) Beckmann(c) Raoult (d) Jean Nollet.

6. In solution, the molecular size of particles is of the order of(a) 10–8 cm (b) 10–9 m(c) > 10–9 m (d) 10–9 cm

7. The new method to determine alcohol content of a wine was discovered in 1870 by(a) Newton (b) Joule(c) Raoult (d) C.A. Wurtz.

8. A solution having three components is called a(a) quaternary solution(b) binary solution(c) single solution(d) ternary solution.

Multiple Choice QuestionsLEVEL - 1


9. An example of a solution having liquid in gas is(a) moist air (b) dry air(c) Au-Hg (d) C2H5OH + H2O

10. Which of the following is not a solution?(a) Air (b) A gold ring(c) Smoke (d) Salt solution

11. Solutions are of ____ types.(a) three (b) six(c) nine (d) eleven

12. Which of the following possesses physical states of solute and solvent as liquid and solid respectively?(a) Solution of sugar in water(b) Zinc amalgam(c) Solution of naphthalene in benzene(d) Brass

13. Which of the following pairs of solutions having different physical states of solute?(a) Homogeneous mixture of chloroform in N2 gas,

solution of CO2 in water(b) Brass, homogeneous mixture of camphor in

N2 gas(c) Sodium amalgam, moist air(d) Solution of H2 gas in Pd metal, mixture of

N2 and O2.

14. In vaporization of liquids into air, liquid dissolves into air hence liquid is(a) solution (b) solute(c) solvent (d) mixture.

15. Sugar dissolved in water is a type of (a) solid in solid solution(b) solid in gas solution(c) solid in liquid solution(d) gas in solid solution.


16. Dissolution of calcium chloride in water is exothermic process while that of ammonium nitrate is endothermic process, the observation is


(a) solubility of calcium chloride decreases with increase in temperature and that of ammonium nitrate increases with increase in temperature

(b) solubilities of both decrease with increase in temperature

(c) solubilities of both increase with increase in temperature

(d) solubility of calcium chloride increases with increase in temperature and that of ammonium nitrate decreases with increase in temperature.

17. 22 cm3 of C2H5OH on dissolving in 200 cm3 of water forms 220 cm3 solution of C2H5OH. What would be the percentage by volume of C2H5OH in the solution?(a) 22% (b) 11%(c) 01% (d) 10%

18. The volume of 2 M NaOH solution is 0.1 dm3. What will be the volume of decimolar NaOH in dm3?(a) 0.1 (b) 0.2(c) 1.5 (d) 2

19. A molal solution is one that contains one mole of a solute in (a) 1000 g of the solvent(b) one litre of the solvent(c) one litre of the solution(d) 1000 g of the solution.

20. When 10 g of caustic soda is dissolved in 250 cm3 of water, molarity of the solution is(a) 1 M (b) 0.5 M(c) 0.25 M (d) 0.1 M

21. A solution contains 0.5 mole of a solute in 400 g of water, its molality would be(a) 1.25 m (b) 0.125 m(c) 2.51 m (d) 0.025 m

22. One part of solute in one million parts of solvent is expressed as(a) ppm (b) milligrams/100 cc(c) grams/litre (d) grams/100 cc.

23. 36 g of glucose (molar mass = 180 g/mol) is present in 500 g of water, the molarity of the solution is(a) 0.2 M (b) 0.4 M(c) 0.8 M (d) 1.0 M

24. The molality of 648 g of pure water is (a) 36 m (b) 55.5 m(c) 3.6 m (d) 5.55 m

25. An aqueous solution contains 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of acetic acid in this solution.

(a) 0.196 (b) 0.301(c) 0.392 (d) 0.503

26. When 0.5 g of solute is present in 107 g of a solution, the ppm of solute will be(a) 0.5 (b) 0.05(c) 106 (d) 107

27. 6 g of urea was dissolved in 500 g of water. The percentage (by mass) of urea in the solution is(a) 0.86% (b) 1.186%(c) 11.86% (d) 0.08%

28. 58 cm3 of ethyl alcohol was dissolved in 400 cm3 of water to form 454 cm3 of solution of ethyl alcohol. The percentage by volume of ethyl alcohol in water is(a) 0.12.% (b) 1.2%(c) 12.78% (d) 0.01%

29. 23 g of ethyl alcohol (molar mass 46 g mol–1) is dissolved in 54 g of water (molar mass 18 g mol–1). The mole fraction of ethyl alcohol and water in solution are respectively(a) 0.7 and 0.3 (b) 0.8571 and 0.1429(c) 0.846 and 0.154 (d) 0.1429 and 0.8571

30. 35% (w/W) solution of ethylene glycol in water, an antifreezer is used in cars as coolant. It lowers freezing point of water to –17.6 °C. The mole fraction of ethylene glycol is(a) 0.1529 (b) 0.1431(c) 0.1343 (d) 0.1634

31. A solution of NaOH (molar mass 40 g mol–1) was prepared by dissolving 1.6 g of NaOH in 500 cm3 of water. The molarity of solution is(a) 0.2 mol dm–3 (b) 0.02 mol dm–3

(c) 0.08 mol dm–3 (d) 0.8 mol dm–3

32. 11.11 g of urea was dissolved in 100 g of water. The molality of solution is (N = 14, H = 1, C = 12, O = 16)(a) 18.52 mol kg–1 (b) 1.852 mol kg–1

(c) 18.52 mol g–1 (d) 1.852 mol g–1

33. 34.2 g of sugar was dissolved in water to produce 214.2 g of sugar syrup. The molality and mole fraction of sugar in the syrup are respectively (C = 12, H = 1, O = 16)(a) 0.224 m, 0.002 (b) 0.05 m, 0.02(c) 0.556 m, 0.0099 (d) 0.05 m, 0.99

34. The molarity and molality of sulphuric acid solution of density 1.198 g cm–3 containing 27% by mass of sulphuric acid (molar mass of H2SO4 = 98 g mol–1) are respectively


(a) 3.301 M, 3.77 m (b) 0.03 M, 0.3 m(c) 0.33 M, 7.3 m (d) 33.1 M, 0.7 m

35. Commercially available concentrated hydrochloric acid is an aqueous solution containing 38% HCl gas by mass. If its density is 1.1 g cm–3, the molarity and mole fraction of HCl solution are respectively(a) 1.14 M and 0.768 (b) 0.114 M and 0.232(c) 11.45 M and 0.232 (d) 11.45 M and 0.768

36. A sample of drinking water was found to be severely contaminated with chloroform which is supposed to be carcinogen. If the level of contamination was 15 ppm (by mass) then the molality of chloroform in the water sample is(a) 1.26 × 10–2 m (b) 1.26 × 10–4 m(c) 12.6 × 10–4 m (d) 0.12 × 10–4 m

37. 34.2 g of glucose is dissolved in 400 g of water. What is the percentage by mass of glucose solution?(a) 78.7% (b) 7.87%(c) 87.7% (d) 77%

38. A solution is prepared by dissolving certain amount of solute in 500 g of water. If the percentage by mass of a solute in solution is 2.38 then the mass of solute will be(a) 1.121 g (b) 121.9 g(c) 12.19 g (d) 1219 g

39. 4.6 cm3 of methyl alcohol is dissolved in 25.2 g of water. The % by mass of methyl alcohol and mole fraction of methyl alcohol are respectively(Given density of methyl alcohol = 0.7952 g cm–3 and C = 12, H = 1, O = 16)(a) 1.26% and 0.755 (b) 12.68% and 0.0755(c) 12% and 0.0075 (d) 1.26% and 0.0755

40. 12.8 cm3 of benzene is dissolved in 16.8 cm3 of xylene. What is the % by volume of benzene?(a) 24.43% (b) 4.324%(c) 43.24% (d) 0.43%

41. What is the mole fraction of HCl in aqueous solution of HCl containing 24.8% of HCl by mass? (H = 1, Cl = 35.5)(a) 1.399 (b) 0.1399(c) 0.0139 (d) 0.0013

42. What is the mole fraction of solute in its 2 molal aqueous solution?(a) 0.0347 (b) 3.47(c) 0.043 (d) 0.074

43. The molality and molarity of HNO3 in a solution containing 12.2% HNO3 are respectively(Given density of HNO3 = 1.038 g cm–3, H = 1, N = 14, O = 16)

(a) 2.01 m and 2.206 M (b) 1.2 m and 2.6 M(c) 2.206 m and 2.01 M (d) 20.6 m and 20 M

44. Sulphuric acid is 95.8% by mass. What is the mole fraction and molarity of H2SO4 solution having density 1.91 g cm–3? (H = 1, S = 32, O = 16)(a) 0.0807 and 1.867 M (b) 0.8073 and 18.67 M(c) 0.07 and 7.86 M (d) 18.67 and 0.8073 M

45. Aqueous solution of NaOH is marked 10% (w/W). The density of the solution is 1.070 g cm–3. The molarity and molality of solution are respectively (Na = 23, H = 1, O = 16)(a) 2.675 M and 2.77 m (b) 0.27 M and 0.26 m(c) 0.26 M and 0.02 m (d) 20.7 M and 20.6 m

46. Battery acid is 4.22 M aqueous H2SO4 solution, and has density of 1.21 g cm–3. What is the molality of H2SO4? (H = 1, S = 32, O = 16)(a) 5.298 mol/g (b) 52.98 mol/kg(c) 5.298 mol/kg (d) 59.98 mol/kg

47. A molal solution is one that contains one mole of solute in(a) one litre of the solvent(b) 1000 g of the solvent(c) one litre of the solution(d) 22.4 litres of solution.

48. Which of the following is independent of temperature?(a) Molarity (b) Molality(c) Both (a) and (b) (d) None of these

49. 20 g of NaOH (molar mass = 40 g mol–1) is dissolved in 500 cm3 of water. Molality of resulting solution is(a) 0.1 m (b) 0.5 m(c) 1.5 m (d) 1.0 m

50. Molarity of solution depends on(a) temperature(b) nature of solute dissolved(c) mass of solvent(d) pressure.

51. Density of water is 1 g/mL. The concentration of water in mol/litre is(a) 1000 (b) 18(c) 0.018 (d) 55.5

52. The solutions A and B are 0.1 M and 0.2 M in a substance. If 100 mL of A is mixed with 25 mL of B and there is no change in volume, the final molarity of the solution is(a) 0.15 M (b) 0.18 M(c) 0.30 M (d) 0.12 M


53. The molarity of a solution that contains 49 g H3PO4 in 2.0 L of a solution is(a) 0.25 M (b) 0.50 M(c) 0.75 M (d) 1.0 M

54. The molarity of the solution containing 7.1 g of Na2SO4 in 100 mL of aqueous solution is(a) 1 M (b) 2 M(c) 0.05 M (d) 0.5 M

55. If 60 cm3 of ethyl alcohol is dissolved in 300 cm3 of water then the percentage by volume of ethyl alcohol is(a) 16.66% (b) 15.76%(c) 17.86% (d) 18.96%

56. What volume of 0.8 M solution contains 0.1 mol of the solute?(a) 62.5 mL (b) 100 mL(c) 500 mL (d) 125 mL

57. 25 mL of 3.0 M HNO3 are mixed with 75 mL of 4.0 M HNO3. If the volumes are additive, the molarity of the final mixture would be(a) 3.25 M (b) 4.0 M(c) 3.75 M (d) 3.50 M

58. The amount of anhydrous Na2CO3 present in 250 mL of 0.25 M solution is(a) 6.225 g (b) 66.25 g(c) 6.0 g (d) 6.625 g

59. 20 mL of HCl solution requires 19.85 mL of 0.01 M NaOH solution for complete neutralization. The molarity of HCl solution is(a) 0.0099 M (b) 0.099 M(c) 0.99 M (d) 9.9 M

60. If 5.85 g of NaCl (molecular weight 58.5 g mol–1) is dissolved in water and the solution is made up to 0.5 litre, the molarity of the solution will be(a) 0.2 M (b) 0.4 M(c) 1.0 M (d) 0.1 M

61. If 1 M and 2.5 litre NaOH solution is mixed with another 0.5 M and 3 litre NaOH solution, then molarity of the resultant solution will be(a) 1.0 M (b) 0.73 M(c) 0.80 M (d) 0.50 M

62. To prepare a solution of concentration of 0.03 g/mL of AgNO3, what amount of AgNO3 should be added in 60 mL of solution?(a) 1.8 g (b) 0.8 g(c) 0.18 g (d) 0.108 g

63. A solution of CaCl2 is 0.5 mol/litre, then the moles of chloride ions in 500 mL will be(a) 0.25 (b) 0.50(c) 0.75 (d) 1.00

64. The sum of mole fractions of A, B and C in an aqueous solution containing 0.2 moles of each A, B and C is(a) 0.6 (b) 0.2(c) 1.0 (d) 3

2.4 Solubility of Gases in Liquids

65. According to Henry's law(a) S = KP (b) S = K/P(c) S = P/K (d) S = KP

66. If Henry's law constant for oxygen is 1.1 × 10–3 mol dm–3 atm–1 and its partial pressure is 0.20 atm, the concentration of dissolved oxygen at NTP will be(a) 2.2 × 10–3 mol dm–3 (b) 2.2 × 10–4 mol dm–3

(c) 1.1 × 10–4 mol dm–3 (d) 0.22 × 10–4 mol dm–3

67. The unit of Henry's constant is(a) mol dm–3 atm–1 (b) mol dm–3 atm(c) mol–1 dm–3 atm (d) mol–1 dm–3 atm–1

68. What is the concentration of dissolved oxygen at 25°C at 1 atmospheric pressure if partial pressure of oxygen is 0.22 atm? The Henry’s law constant for oxygen is 1.3 × 10–3 mol dm–3 atm–1.(a) 1.3 × 10–3 mol dm–3 (b) 0.13 × 10–3 mol dm–3

(c) 0.28 × 10–4 mol dm–3 (d) 2.86 × 10–4 mol dm–3

69. What is the Henry’s law constant of dissolved O2 at 10°C at 1 atmospheric pressure, if partial pressure of oxygen is 0.24 atm? The concentration of dissolved oxygen is 3.12 × 10–4 mol dm–3.(a) 2.5 × 10–3 mol dm–3 atm–1

(b) 1 × 10–4 mol dm–3 atm–1

(c) 1.3 × 10–3 mol dm–3 atm–1

(d) data insufficient.

70. Solubility of a gas in a liquid increases with(a) increase of pressure and increase of temperature(b) decrease of pressure and increase of temperature(c) increase of pressure and decrease of temperature(d) decrease of pressure and decrease of temperature.

71. Which law states that the amount of gas dissolved in a given mass of solvent at any temperature is directly proportional to pressure of the gas above the solution?(a) Raoult’s law (b) Boyle’s law(c) Charles’ law (d) Henry’s law


2.5 Solid Solutions

72. Lead is hardened by the addition of (a) 10-20% aluminium (b) 5-20% manganese(c) 10-20% antimony (d) 20-30% iron

73. Babbitt metal is an alloy of (a) bismuth with tin and copper(b) antimony with tin and copper(c) lead with arsenic (d) aluminium with copper and manganese.

74. Amalgam is(a) a solution of gas in gas(b) a solution of liquid in gas(c) a solution of metals in liquid metal(d) a solution of solid in gas.

75. Stainless steel contains(a) chromium (b) nickel(c) both (a) and (b) (d) none of these.

76. An alloys containing zero temperature coefficient of electrical resistance is(a) manganin (b) duralumin(c) bronze (d) spiegeleisen.

77. An alloy is(a) a solution of solid in liquid(b) a solution of solid in gas(c) a solution of solid in plasma(d) a solution of solid in solid.

78. 5 to 20% manganese in iron is(a) ferromanganeous (b) manganin(c) duralumin (d) spiegeleisen.

79. An alloy manganin contains (a) 84% Cu, 12% Mn and 4% Ni(b) 70-80% Mn, 30-20% Fe(c) 10-20% Sb, 90-80% Pb(d) Al, Cu, Mg, and Mn.

2.6 Colligative Properties

80. Which of the following is not a colligative property?(a) Osmotic pressure(b) Elevation in boiling point(c) Vapour pressure(d) Depression in freezing point

81. Which of the following is a colligative property?(a) Osmotic pressure (b) Boiling point(c) Vapour pressure (d) Freezing point

82. Colligative properties are applicable to(a) ideal dilute solutions(b) non-ideal concentrated solutions(c) non-ideal solutions(d) ideal concentrated solutions.

83. The colligative properties of a solutions depend on(a) nature of solute particles present in it(b) nature of solvent used(c) number of solute particles present in it(d) number of moles of solvent only.

84. Which of the following is not the colligative property?(a) DTf (b) DTb

(c) Kb (d) Osmotic pressure

85. Colligative properties are used for the determination of(a) molar mass(b) equivalent weight(c) arrangement of molecules(d) melting point and boiling point.

86. Which of the following is a colligative property?(a) Conductance of a solution(b) Surface tension of a solution(c) Osmotic pressure of a solution(d) Radioactivity of a solution

87. Which is not a colligative property?(a) Refractive index(b) Lowering of vapour pressure(c) Depression of freezing point(d) Elevation of boiling point

2.7 Lowering of Vapour Pressure

88. Which of the following is incorrect?(a) Relative lowering of vapour pressure is

independent of the nature of the solute and the solvent

(b) Relative lowering in vapour pressure is a colligative property

(c) Vapour pressure of a solution is lower than the vapour pressure of the solvent

(d) Relative lowering of vapour pressure is directly proportional to the original pressure.

89. Which one of the following is not an expression for Raoult’s law for a solution containing two volatile components A and B? pA and pB are the partial vapour pressures of A and B, p°A and p°B are the vapour pressures of pure A and B, xA and xB are mole fractions of A and B in solution.


(a) pA = p°A xA (b) Ptotal = p°AxA + p°BxB

(c) Dp = p°BxB (d) Dp = p°AxA

90. Lowering of vapour pressure is a colligative property because, it depends on the concentration of(a) volatile solute (b) non-volatile solute(c) volatile solvent (d) non-volatile solvent.

91. At 25°C, the total pressure of an ideal solution obtained by mixing 3 moles of A and 2 moles of B, is 184 torr. What is the vapour pressure (in torr) of pure B at the same temperature? (vapour pressure of pure A, at 25°C, is 200 torr.)(a) 180 (b) 160(c) 16 (d) 100

92. The relative lowering of vapour pressure is equal to

(a) pp°D

1 (b) p pp− °°

1

1

(c) D°p

p1

(d) p pp

° −D

1

93. The aqueous solution that has the highest value of relative lowering of vapour pressure at a given temperature is(a) 0.1 molal sodium phosphate(b) 0.1 molal barium chloride(c) 0.1 molal sodium chloride(d) 0.1 molal glucose.

94. For dilute solutions, Raoult’s law states that(a) lowering of vapour pressure is equal to the mole

fraction of the solute(b) relative lowering of vapour pressure is equal to

the mole fraction of the solvent(c) relative lowering of vapour pressure of the

solvent is equal to the mole fraction of the solute

(d) vapour pressure of the solution is equal to the vapour pressure of the solvent.

95. Vapour pressure of dilute aqueous solution of glucose is 750 mm of mercury at 373 K. The mole fraction of solute is (a) 1/76 (b) 1/7.6(c) 1/38 (d) 1/10

96. Vapour pressure of a solution of 5 g of non-electrolyte in 100 g of water at a particular temperature is 2985 N/m2. The vapour pressure of pure water is 3000 N/m2, the molecular weight of the solute is (a) 90 g mol–1 (b) 200 g mol–1

(c) 180 g mol–1 (d) 380 g mol–1

97. The vapour pressure of a solution containing 13 × 10–3 kg of solute in 0.1 kg of water at 298 K is 27.371 mm Hg. Given that the vapour pressure of water at 298 K is 28.065 mm Hg. The molar mass of the solute will be(a) 69 g mol–1 (b) 95 g mol–1

(c) 60 g mol–1 (d) 90 g mol–1

98. A solution is prepared from 26.2 × 10–3 kg of an unknown substance and 112.0 × 10–3 kg acetone at 313 K. The vapour pressure of pure acetone at this temperature is 0.526 atm. The vapour pressure of solution is 273.52 × 10–3 kg mol–1 if the molar mass of substance is(a) 0.400 atm (b) 0.600 atm(c) 0.500 atm (d) 0.300 atm

99. The vapour pressure of 2.1% solution of a non-electrolyte in water at 100°C is 755 mm Hg. What is the molar mass of the solute?(a) 58.69 kg/mol (b) 58.69 g/mol(c) 5.86 kg/mol (d) 586.9 g/mol

100. The vapour pressure of water at 20°C is 17 mm Hg. What is the vapour pressure of a solution containing 2.8 g of urea (NH2CONH2) in 50 g of water? (N = 14, C = 12, H = 1)(a) 16.71 mm of Hg (b) 1.671 mm of Hg(c) 0.1671 mm of Hg (d) 0.01671 mm of Hg

101. In an experiment, 18.04 g of mannitol was dissolved in 100 g of water. The vapour pressure of water was lowered by 0.309 mm Hg from 17.535 mm Hg. The molar mass of mannitol is(a) 18.42 g/mol (b) 18.42 kg/mol(c) 1.842 g/mol (d) 184.27 g/mol

102. According to the Raoult’s law, the relative lowering of vapour pressure is equal to the(a) mole fraction of solvent(b) mole fraction of solute(c) independent of mole fraction of solute(d) molality of solution.

103. Partial pressure of solvent in solution of non-volatile solute is given by equation,(a) p = x2p1° (b) p1° = xp(c) p = x1p1° (d) p1° = x1p

104. When partial pressure of solvent in solution of non-volatile solute is plotted against its mole fraction, nature of graph is(a) a straight line passing through origin(b) a straight line parallel to mole fraction of

solvent


(c) a straight line parallel to vapour pressure of solvent

(d) a straight line intersecting vapour pressure axis.

105. Relative lowering of vapour pressure of solution is a(a) property of solute(b) property of solute as well as solvent(c) property of solvent(d) colligative property.

106. Vapour pressure of solution of a non-volatile solute is always(a) equal to the vapour pressure of pure solvent(b) higher than vapour pressure of pure solvent(c) lower than vapour pressure of pure solvent(d) constant.

107. The vapour pressure at equilibrium of a liquid in a closed vessel depends on(a) pressure (b) concentration(c) temperature (d) volume.

108. The vapour pressure of water at 300 K in a closed container is 0.4 atm. If the volume of the container is doubled, its vapour pressure at 300 K will be (a) 0.8 atm (b) 0.2 atm(c) 0.4 atm (d) 0.6 atm

109. At 300 K when a solute is added to a solvent its vapour pressure over the mercury reduces from 50 mm to 45 mm. The value of mole fraction of solute will be(a) 0.005 (b) 0.010(c) 0.100 (d) 0.900

110. If p1° and p are the vapour pressures of a solvent and its solution respectively, and if x1 and x2 are the mole fractions of the solvent and solute respectively then(a) p = p1° x1 (b) p = p1° x2

(c) p1° = p x2 (d) p = p1° (x1/x2)

111. The vapour pressure of pure liquid A is 0.80 atm. On mixing a non-volatile solute B to A, its vapour pressure becomes 0.6 atm. The mole fraction of B in the solution is(a) 0.150 (b) 0.25(c) 0.50 (d) 0.75

112. The vapour pressure of two liquids P and Q are 80 and 60 torr respectively. The total vapour pressure of solution obtained by mixing 3 moles of P and 2 moles of Q would be(a) 140 torr (b) 20 torr(c) 68 torr (d) 72 torr

113. 60 g of urea (mol. wt. 60) was dissolved in 9.9 moles of water. If the vapour pressure of pure water is p1° , the vapour pressure of solution is(a) 0.10 p1° (b) 1.10 p1° (c) 0.90 p1° (d) 0.99 p1°

114. The vapour pressure of a solvent is decreased by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if the decrease in the vapour pressure is 20 mm of mercury?(a) 0.8 (b) 0.6(c) 0.4 (d) 0.2

115. The vapour pressure of a pure liquid A is 70 torr at 27°C. It forms an ideal solution with another liquid B. The mole fraction of B is 0.2 and the total vapour pressure of the solution is 84 torr at 27°C. The vapour pressure of pure liquid B at 27°C is(a) 14 torr (b) 56 torr(c) 70 torr (d) 140 torr

116. At 40°C the vapour pressure of pure liquids, benzene and toluene, are 160 mm Hg and 60 mm Hg respectively. At the same temperature, the vapour pressure of an equimolar solution of the two liquids, assuming the ideal solution should be(a) 140 mm Hg (b) 110 mm Hg(c) 220 mm Hg (d) 100 mm Hg

117. One mole of sugar is dissolved in two moles of water. The vapour pressure of the solution relative to that of pure water is(a) 2/3 (b) 1/3(c) 3/2 (d) 1/2

118. The vapour pressure of benzene at 80°C is lowered by 10 mm by dissolving 2 g of a non-volatile substance in 78 g of benzene. The vapour pressure of pure benzene at 80°C is 750 mm. The molecular mass of the substance will be(a) 15 g mol–1 (b) 150 g mol–1

(c) 1500 g mol–1 (d) 148 g mol–1

119. Vapour pressure of CCl4 at 25°C is 143 mm of Hg. 0.5 g of a non-volatile solute (mol. wt. = 65) is dissolved in 100 mL CCl4. Find the vapour pressure of the solution. (Density of CCl4 = 1.58 g/cm3)(a) 141.43 mm (b) 94.39 mm(c) 199.34 mm (d) 143.99 mm

120. The vapour pressure of a solution increases when(a) the temperature is raised(b) the volume is increased(c) the number of moles of the solute is increased(d) the temperature is lowered.


121. Pressure cooker reduces cooking time for food because(a) heat is more evenly distributed in the cooking

space(b) boiling point of water involved in cooking is

increased(c) the higher pressure inside the cooker crushed

the food material(d) cooking involves chemical changes helped by a

rise in temperature.

122. The relative lowering of the vapour pressure is equal to the ratio between the number of(a) solute molecules and solvent molecules(b) solute molecules and the total molecules in the

solution(c) solvent molecules and the total molecules in the

solution(d) solvent molecules and the total number of ions

of the solute.

123. The vapour pressure lowering caused by the addition of 100 g of sucrose (molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at 25°C is 23.8 mm Hg is(a) 1.25 mm Hg (b) 0.125 mm Hg(c) 1.15 mm Hg (d) 00.12 mm Hg

2.8 Boiling Point Elevation

124. Boiling point of water is 373.11 K. What is Kb of water, if 0.15 molal aqueous solution of a substance boils at 373.20 K?(a) 0.09 K kg mol–1 (b) 0.6 K kg mol–1

(c) 6.0 K kg mol–1 (d) 60 K kg mol–1

125. Kb is the elevation in boiling point produced when 1 mol of solute is dissolved in(a) one litre solvent (b) 1000 g solvent(c) 500 mL solvent (d) 0.5 kg solvent.

126. If 0.15 molal solution of a substance boils at 373.23 K the molal elevation constant of water is (Given boiling point of water = 373.15 K)(a) 0.35 K kg mol–1 (b) 5.3 K kg mol–1

(c) 0.53 K kg mol–1 (d) 0.053 K kg mol–1

127. A solution of a substance on dissolving in benzene boils at 353.71 K. If Kb for benzene is 2.53 K kg mol–1 and boiling point of pure benzene is 353.35 K, then the molality of the solution is(a) 0.1423 m (b) 0.4123 m(c) 0.04 m (d) 0.004 m

128. A solution is prepared by dissolving 1.9 × 10–2 kg ammonia in 400 g of water. The elevation in boiling point if Kb for water is 0.52 K kg mol–1 (N = 14, H = 1, O = 16) is(a) 145°C (b) 1.45 K(c) 14.5 K (d) 45.1 K

129. A solution containing 1.21 g of camphor (molar mass 152 g mol–1) in 26.68 g of acetone boils at 329.95 K. If the boiling point of pure acetone is 329.45 K, the molal elevation constant for acetone will be(a) 167 K kg mol–1 (b) 1.67 K mol–1

(c) 16.76 K kg mol–1 (d) 1.676 K kg mol–1

130. 3.795 g of sulphur is dissolved in 100 g of CS2. This solution boils at 319.81 K. What is the molecular formula of sulphur in solution? The boiling point of CS2 is 319.45 K. (Given : Kb for CS2 = 2.42 K kg mol–1 and atomic mass of S = 32.)(a) S2 (b) S4

(c) S8 (d) None of these.

131. A solution prepared by dissolving certain amount of compound in 31.8 g of CCl4 has a boiling point of 0.392 K higher than that of pure CCl4. If the molar mass of compound is 128 g mol–1, the mass of the solute dissolved will be(Given Kb for CCl4 = 5.02 K kg mol–1)(a) 317 g (b) 0.317 kg(c) 0.317 g (d) 31.7 g

132. Boiling point of water at 750 mm Hg is 96.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C? Molal elevation constant for water is 0.52 K kg mol–1.(a) 11.08 kg (b) 1108.2 g(c) 1108.2 kg (d) 11.08 g

133. A solution containing 0.5126 g of naphthalene (molar mass = 128.17 g mol–1) in 50.00 g of CCl4 gives a boiling point elevation of 0.402 K. While a solution of 0.6216 g of unknown solute in the same mass of the solvent gives a boiling point elevation of 0.647 K. The molar mass of the unknown solute is (Kb for CCl4 = 5.03 K kg mol–1)(a) 0.096306 kg/mol (b) 96.31 kg/mol(c) 0.009630 kg/mol (d) 0.9631 kg/mol

134. The mass in grams of an impurity of molar mass 100 g mol–1 which would be required to raise the boiling point of 50 g of chloroform by 0.30 K (Kb for chloroform = 3.63 K kg mol–1) is(a) 0.4132 g (b) 41.32 g(c) 43.1 g (d) 42.3 g


135. Boiling point of a solvent is 80.2°C. When 0.419 g of the solute of molar mass 252.4 g mol–1 was dissolved in 75 g of the solvent, the boiling point of the solution was found to be 80.256°C. What is the molal elevation constant ?(a) 25.4 K kg/mol (b) 2.54 K g/mol(c) 25.4 K g/mol (d) 2.54 K kg/mol

136. When NaCl is added to water(a) freezing point is raised(b) boiling point is depressed(c) freezing point does not change(d) boiling point is raised.

137. Molal elevation constant is elevation in boiling point produced by(a) 1 g of solute in 100 g of solvent(b) 100 g of solute in 1000 g of solvent(c) 1 mole of solute in one litre of solvent(d) 1 mole of solute in one kg of solvent.

138. The determination of molar mass from elevation in boiling point is called as(a) cryoscopy (b) osmometry(c) ebullioscopy (d) spectroscopy.

139. If mass is expressed in gram then Kb is given by

(a) M T WW

b2 2

11000D ×

×

(b) W

T W Mb

2

1 21000

D × ××

(c) M T WW

b2 1

21000D ×

×

(d) W

T W Mb

1

2 21000

D × ××

140. Unit of boiling point elevation constant (Kb) is(a) kg mol–1 (b) K mol–1

(c) g mol–1 (d) K kg mol–1

141. The rise in the boiling point of a solution containing 1.8 g of glucose in 100 g of a solvent is 0.1°C. The molal elevation constant of the liquid is(a) 0.01 K/m (b) 0.1 K/m(c) 1 K/m (d) 10 K/m

142. The molal elevation of boiling point constant for water is 0.513°C kg mol–1. When 0.1 mole of sugar is dissolved in 200 mL of water, the solution boils under a pressure of one atmosphere at(a) 100.513 °C (b) 100.0513 °C(c) 100.256 °C (d) 101.025 °C

143. An aqueous solution containing 1 g of urea boils at 100.25°C. The aqueous solution containing 3 g of glucose in the same volume will boil at (molecular weights of urea and glucose are 60 and 180 respectively)(a) 100.75 °C (b) 100.5 °C(c) 100.25 °C (d) 100 °C

144. If 0.15 g of a solute dissolved in 15 g of solvent is boiled at a temperature higher by 0.216 °C than that of the pure solvent, the molecular weight of the substance (molal elevation of boiling point constant for the solvent is 2.16 °C) is(a) 1.01 g mol–1 (b) 10 g mol–1

(c) 1000 g mol–1 (d) 100 g mol–1

145. A liquid is in equilibrium with its vapours at its boiling point. On average, the molecules in the two phases have equal(a) intermolecular forces(b) potential energy(c) total energy(d) kinetic energy.

146. Which of the following aqueous solutions containing 10 g of solute in each case has the highest boiling point?(a) NaCl solution (b) KCl solution(c) Sugar solution (d) Glucose solution

147. Elevation in boiling point was 0.52°C when 6 g of a compound X was dissolved in 100 g of water. Molecular weight of X is (Kb for water is 0.52 K kg mol–1).(a) 120 g mol–1 (b) 60 g mol–1

(c) 180 g mol–1 (d) 600 g mol–1

148. At higher altitudes the boiling point of water is lower because(a) atmospheric pressure is low(b) temperature is low(c) atmospheric pressure is high(d) temperature is high.

149. The boiling point of C6H6, CH3OH, C6H5NH2 and C6H5NO2 are 80°C, 65°C, 184°C and 212°C respectively. Which will show highest vapour pressure at room temperature?(a) C6H6 (b) CH3OH(c) C6H5NH2 (d) C6H5NO2

150. The boiling point of 0.1 molal aqueous solution of urea is 100.18°C at 1 atm. The molal elevation constant of water (in °C kg mol–1) is(a) 1.8 (b) 0.18(c) 18 (d) 18.6


151. If the solution boils at a temperature T1 and the solvent at a temperature T2, the elevation of boiling point is given by(a) T1 + T2 (b) T1 – T2(c) T2 – T1 (d) T1 ÷ T2

152. The molal elevation constant is the ratio of the elevation in boiling point to(a) molarity(b) molality(c) mole fraction of solute(d) mole fraction of solvent.

153. Elevation of boiling point was 0.52°C, when 6 g of compound ‘X’ was dissolved in 100 g of H2O. Molar mass of ‘X’ is (Kb of water is 5.2°C per 100 g water)(a) 120 g mol–1 (b) 60 g mol–1

(c) 600 g mol–1 (d) 180 g mol–1

154. The elevation of boiling point method is used for the determination of molecular weight of(a) non-volatile and soluble solute(b) non-volatile and insoluble solute(c) volatile and soluble solute(d) volatile and insoluble solute.

155. When a substance is dissolved in a solvent, the vapour pressure of the solvent is decreased. This results in(a) an increase in the boiling point of the solution(b) a decrease in the boiling point of the solvent(c) the solution having a higher freezing point than

the solvent(d) the solution having a lower osmotic pressure

than the solvent.

156. When 10 g of a non-volatile solute is dissolved in 100 g of benzene, it raises boiling point by 1°C, then molecular mass of the solute is (Kb for benzene = 2.53 K kg mol–1)(a) 223 g (b) 233 g(c) 243 g (d) 253 g

157. The value of Kb for a solvent is X K kg mol–1. A 0.2 m solution of a non-electrolyte in this solvent will boil at (Given : boiling point of solvent = A °C).

(a) (A + X) °C (b) A X+

°

10C

(c) A X+

°

5C (d) A X+

5

K

158. At certain hill-station pure water boils at 99.725°C. If Kb for water is 0.513 °C kg mol–1, the boiling point of 0.69 m solution of urea will be(a) 100.079°C (b) 103°C(c) 100.359°C (d) Unpredictable


159. Which of the following solutions shows maximum depression in freezing point?(a) 0.5 M Li2SO4 (b) 1 M NaCl(c) 0.5 M Al2(SO4)3 (d) 0.5 M BaCl2

160. The unit of freezing point depression constant is(a) K mol–1 (b) K kg–1 mol–1

(c) K kg mol–1 (d) K kg–1

161. The freezing point of equimolal aqueous solution will be highest for (a) C6H5NH3

+Cl– (b) La(NO3)3

(c) glucose (d) Ca(NO3)2

162. A 0.5 molal solution of ethylene glycol in water is used as coolant in a car. If Kf for water is 1.86 K kg mol–1

the mixture will freeze at(a) 0.93°C (b) –0.93°C(c) 1.86°C (d) –1.86°C

163. 5.08 g of a substance dissolved in 50 g of water lowered the freezing point by 1.2 K. If the molal depression constant for water is 1.86 K kg mol–1, the molar mass of the substance, will be(a) 157.48 × 10–3 g mol–1 (b) 157.48 g mol–1

(c) 15.7 × 10–3 g mol–1 (d) 15.7 kg mol–1

164. 0.440 × 10–3 kg of a substance (molar mass = 178.9 × 10–3 kg mol–1) dissolved in 22.2 × 10–3 kg of benzene lowered the freezing point of benzene by 0.567 K. The molal depression constant for benzene is(a) 512 K kg mol–1 (b) 5.12 K kg mol–1

(c) 51.2 K kg mol–1 (d) none of these.

165. The observed depression in the freezing point of water for a particular solution is 0.087 K. The molality of the solution if molal depression constant for water is 1.86 K kg mol–1 will be(a) 0.4 m (b) 4.67 m(c) 0.0467 m (d) 4 m

166. 1.02 g of urea when dissolved in 98.5 g of certain solvent decreases its freezing point by 0.211 K. 1.60 g of unknown compound when dissolved in 86.0 g of the same solvent depresses the freezing point by 0.34 K. The molar mass of the unknown compound is (Urea NH2CONH2, N = 14, C = 12, O = 16, H = 1)(a) 60 g mol–1 (b) 68 g mol–1

(c) 78 g mol–1 (d) 40 g mol–1


167. A solution of glucose (C6H12O6) was prepared by dissolving certain amount of glucose in 100 g of water. The depression in freezing point was 0.0410 K. If molal depression constant for water is 1.86 K kg mol–1, the mass of glucose dissolved will be (C = 12, H = 1, O = 16)(a) 0.396 g (b) 0.396 kg(c) 39.6 g (d) 39.6 kg

168. An aqueous solution containing 12.5 × 10–3 kg of non-volatile compound in 0.1 kg of water freezes at 272.49 K. The molar mass of the compound is (Kf for water = 1.86 K kg mol–1.(a) 35.27 g mol–1 (b) 352.27 g mol–1

(c) 452.27 g mol–1 (d) 45.27 g mol–1

169. The freezing point of solution prepared by dissolving 4.5 g of glucose (Molar mass = 180 g mol–1) in 250 g of bromoform is (Given : Freezing point of bromoform = 7.8°C and Kf for bromoform = 14.4 K kg mol–1)(a) 6.36°C (b) 4.23°C(c) 3.66°C (d) 0.06°C

170. A temperature at which the vapour pressure of a solid is equal to the vapour pressure of liquid is called(a) elevation of boiling point(b) freezing point(c) melting point(d) depression of freezing point

171. The freezing point of 1 percent solution of lead nitrate in water will be(a) below 0 °C (b) 0 °C(c) 1 °C (d) 2 °C

172. How much polystyrene of molecular weight 9000 would have to be dissolved in 100 grams of C6H6 to lower its freezing point by 1.05°C? (Kf C6H6 = 4.9)(a) 19.3 g (b) 193 g(c) 38.6 g (d) 77.2 g

173. Relationship between Kf , m and DTf can be written as(a) DTf = Kf /m (b) DTf = Kf m(c) DTf = Kf (d) DTf = m

174. Which will show maximum depression in freezing point when concentration is 0.1 M?(a) NaCl (b) Urea(c) Glucose (d) K2SO4

175. In cold countries, ethylene glycol is added to water in the radiators of cars during winter. It results in

(a) reducing viscosity(b) reducing specific heat(c) reducing freezing point(d) reducing boiling point.

176. The molar mass of the solute using depression of freezing point may be calculated using

(a) MK W

T mf

f2

210=

D (b) M

K W

T Wf

f2

2

1

100=

D

(c) MK W

T Wf

f2

1

2=D

(d) MK W

T Wf

f2

2

1

1000=

D

177. The freezing point of 0.05 molal solution of non-electrolyte in water is (Kb = 1.86 °C kg mol–1)(a) –1.86 °C (b) –0.93 °C(c) –0.093 °C (d) 0.93 °C

178. What is the molality of solution of a certain solute in a solvent, if there is a freezing point depression of 0.184 °C, and the freezing point constant is 18.4?(a) 0.01 m (b) 1 m(c) 0.001 m (d) 100 m

179. If all the following four compounds were sold at the same price, which would be cheapest for preparing an antifreeze solution for a car radiator?(a) Methanol (b) Ethanol(c) Ethylene glycol (d) Glycerol

180. Equimolal solutions of A and B show depression of freezing point in the ratio of 2 : 1. A remains in normal state. The state of B in the solution is(a) normal (b) dissociated(c) associated (d) unpredictable.

181. A solution containing 6.8 g of a non-ionic solute in 100 g water was found to freeze at –0.93 °C. The freezing point depression constant of water is 1.86. The molar mass of the solute is(a) 13.6 g mol–1 (b) 34 g mol–1

(c) 68 g mol–1 (d) 136 g mol–1

182. Which of the following aqueous solutions has minimum freezing point?(a) 0.01 m NaCl (b) 0.005 m C2H5OH(c) 0.005 m MgI2 (d) 0.001 m MgSO4

183. When mercuric iodide is added to the aqueous solution of potassium iodide, the(a) freezing point is lowered(b) freezing point is raised(c) freezing point does not change(d) boiling point does not change.


184. An aqueous solution freezes at 0.186°C. The boiling point of the same solution is(Kf = 1.86 K m–1, Kb = 0.512 K m–1)(a) 100.186 °C (b) 100.512 °C

(c) 100 5120 186

..

(d) 100.0512 °C

185. An aqueous solution of a non-electrolyte boils at 100.52°C. The freezing point of the solution will be (Kf = 1.86 K m–1, Kb = 0.512 K m–1)(a) 0°C (b) –1.86°C(c) 1.86°C (d) –0.52°C

186. A solution of urea (mol. mass 56 g mol–1) boils at 100.18° C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol–1 respectively the above solution will freeze at(a) –6.54°C (b) 6.54°C(c) 0.654°C (d) –0.654°C

187. After adding a solute freezing point of solution decreases to –0.186. Calculate DTb if Kf = 1.86 and Kb = 0.521.(a) 0.521 (b) 0.0521(c) 1.86 (d) 0.0186

188. Which of the following will have the highest freezing point at one atmosphere?(a) 0.1 M NaCl solution (b) 0.1 M sugar solution(c) 0.1 M BaCl2 solution (d) 0.1 M FeCl3 solution

189. 6 g of substance x dissolved in 100 g of water freezes at –0.93 °C. The molar mass of x is (Kf = 1.86 K m–1)(a) 60 g mol–1 (b) 120 g mol–1

(c) 180 g mol–1 (d) 140 g mol–1

190. What is the molality of ethyl alcohol (mol. wt. = 46) in aqueous solution which freezes at –10°C? (Kf for water = 1.86 K molality–1)(a) 3.540 m (b) 4.567 m(c) 5.376 m (d) 6.315 m

191. If Kf value of H2O is 1.86, the value of DTf for 0.1 m solution of non-volatile solute is(a) 18.6 (b) 0.186(c) 1.86 (d) 0.0186

192. 1% solution of Ca(NO3)2 has the freezing point(a) 0°C (b) less than 0°C(c) greater than 0°C (d) 273 K

193. If a substance exists as trimer in the solution, then which of the following alternative is possible for depression in freezing point of m molal solution?

(a) mK f

3 (b)

mK f

4

(c) mK f

5 (d)

mK f

8

2.10 Osmosis and Osmotic Pressure

194. 0.6 g of a solute is dissolved in 0.1 L of a solvent which develops an osmotic pressure of 1.23 atm at 27°C. The molecular mass of the substance is(a) 149.5 g mol–1 (b) 120 g mol–1

(c) 430 g mol–1 (d) 102 g mol–1

195. Solutions having same osmotic pressure are known as(a) hypotonic (b) hypertonic(c) isotonic (d) isomeric.

196. The osmotic pressure of a dilute solution is directly proportional to the (a) diffusion rate of the solute(b) concentration(c) boiling point(d) flow of solvent from a concentrated to a dilute

solution.

197. At a given temperature isotonic solutions have the same(a) density(b) volume(c) normality(d) molar concentration.

198. The osmotic pressure is expressed in units of(a) MeV (b) calories(c) cm (d) atmosphere.

199. The scientist, who discovered the phenomenon of osmosis in natural membranes is(a) Raoult F. Marie (b) E. Otto(c) van't Hoff (d) Abbe Nollet.

200. What happens when blood cells are placed in pure water?(a) The fluid in blood cells rapidly moves into

water(b) The water molecules rapidly move into blood

cells(c) The blood cells dissolve in water(d) No change takes place.

201. Calculate the osmotic pressure of a solution containing 0.153 g glucose in 0.1 litre of the solution at 298 K?(a) 0.027 atm (b) 0.107 atm(c) 0.172 atm (d) 0.207 atm

202. The osmotic pressure of 4.5 g of glucose (molar mass = 180 g mol–1) dissolved in 100 mL of water at 298 K (R = 0.0821 L atm mol–1 K–1) is(a) 5.116 atm (b) 61.6 atm(c) 6.116 atm (d) 611.6 atm


203. A solution containing 17.8 g L–1 of cane sugar (molar mass 342 g mol–1) has an osmotic pressure 1.2 atm. What is the temperature of the solution?

(R = 0.082 L atm mol–1 K–1).(a) 281.4°C (b) 428.1 K(c) 281.4 K (d) 428.1°C

204. 30 g of glucose dissolved in one litre of water has an osmotic pressure 4.91 atm at 303 K. If the osmotic pressure of the glucose solution is 1.5 atm at the same temperature, what would be its concentration? (Molar mass of glucose is 180 g mol–1)(a) 0.5 M (b) 0.0509 M(c) 5 M (d) None of these

205. A solution has an osmotic pressure of 3.90 × 105 Nm–2 at 300 K. The volume containing 1 mole of solute if solution of same solute has an osmotic pressure of 2.82 × 105 Nm–2 and contains 1 mole of solute in 10.5 m3 is(a) 7.59 m3 (b) 7.59 L(c) 7.59 cm3 (d) 75.9 m3

206. What is the mass of sucrose in its 1 L solution (molar mass = 342 g mol–1) which is isotonic with 6.6 × 10–3 kg L–1 of urea (NH2CONH2)?(Given : atomic masses H = 1, C = 12, N = 14, O = 16 in g mol–1)(a) 37.62 kg (b) 3.762 g(c) 37.62 g (d) 376.2 g

207. Osmotic pressure of a solution containing 6.8 × 10–3 kg of protein per 1 × 10–4 m3 of solution is 3.02 × 103 Pa at 37°C. What is the molar mass of protein? (R = 8.314 J K–1 mol–1)(a) 55.06 kg mol–1 (b) 58.06 kg mol–1

(c) 68.05 kg mol–1 (d) 65.08 kg mol–1

208. At 298 K, 1000 cm3 of a solution containing 4.34 g of solute shows osmotic pressure of 2.55 atm. What is the molar mass of solute?

(R = 0.0821 L atm K–1 mol–1)(a) 44.16 g mol–1 (b) 41.46 g mol–1

(c) 41.64 g mol–1 (d) 46.14 g mol–1

209. What is the volume of a solution containing 34.2 g of cane sugar (molar mass = 342 g mol–1) which has an osmotic pressure 2.42 atm at 20°C?(a) 0.9945 L (b) 0.9945 × 10–3 L(c) 0.09945 L (d) 0.09945 × 10–3 L

210. A solution of particular amount of organic substance of molar mass 196 g mol–1 dissolved in 2 litres of water gave an osmotic pressure of 0.54 atm at 12°C. The mass of the solute dissolved is

[R = 0.0821 L atm K–1 mol–1]

(a) 10.5 g (b) 9.04 kg(c) 9.04 g (d) 0.9 kg

211. Which of the following 0.1 M aqueous solutions will exert highest osmotic pressure?(a) NaCl (b) BaCl2(c) MgSO4 (d) Al2(SO4)3

212. In osmosis(a) solvent molecules pass from high concentration

of solute to low concentration.(b) solvent molecules pass from a solution of low

concentration of solute to a solution of high concentration of solute.

(c) solute molecules pass from low concentration to high concentration

(d) solute molecules pass from high concentration to low concentration.

213. A solution contains non-volatile solute of molecular mass MB. Which of the following can be used to calculate molecular mass of the solute in terms of osmotic pressure? (WB = Mass of solute, V = Volume of solution and p = Osmotic pressure)

(a) MB = W

VRTBp

(b) MB = W RT

VBp

(c) MB = W V

RTBp

(d) MB = WV

RTB p

214. A membrane which allows solvent molecules but not the solute molecules to pass through it is called as(a) semipermeable membrane(b) permeable membrane(c) filter membrane(d) porous membrane.

215. Which inorganic precipitate acts as semipermeable membrane?(a) Calcium sulphate (b) Barium oxalate(c) Nickel phosphate (d) Copper ferrocyanide

216. The osmotic pressure of a solution increases if(a) temperature is decreased(b) solution constant is increased(c) number of solute molecule is increased(d) volume is increased.

217. The solution in which the blood cells retain their normal form are _______ to the blood.(a) isotonic (b) isomotic(c) hypertonic (d) equinormal

218. In equimolar solutions of glucose, NaCl and BaCl2, the order of osmotic pressure is(a) Glucose > NaCl > BaCl2(b) NaCl > BaCl2 > Glucose


(c) BaCl2 > NaCl > Glucose(d) Glucose > BaCl2 > NaCl

219. A solution having a higher osmotic pressure than another solution, is called a(a) hypotonic solution (b) isotopic solution(c) isotonic solution (d) hypertonic solution.

220. Water transportation in plants takes place by the phenomenon of(a) diffusion (b) osmosis(c) reverse osmosis (d) reverse diffusion.

221. The solution containing 4.0 g of a polyvinyl chloride polymer in 1 litre of dioxane was found to have an osmotic pressure 6.0 × 10–4 atmosphere at 300 K, the value of R used is 0.082 L atm mol–1 K–1. The molecular mass of the polymer was found to be(a) 3.0 × 102 (b) 1.6 × 105

(c) 5.6 × 104 (d) 6.4 × 102

222. If mole fraction of the solvent in a solution decreases then(a) vapour pressure of solution increases(b) boiling point decreases(c) osmotic pressure increases(d) none of these.

223. The concentration in g/L of a solution of cane sugar (molar mass = 342 g mol–1) which is isotonic with a solution containing 6 g of urea (molar mass = 60 g mol–1) per litre is(a) 3.42 (b) 34.2(c) 5.7 (d) 19

224. The osmotic pressure of 10% solution of cane sugar at 69°C in atmospheres is(a) 724 (b) 824(c) 8.21 (d) 7.21

225. The relationship between osmotic pressure at 273 K when 10 g glucose (P1), 10 g urea (P2) and 10 g sucrose (P3) are dissolved in 250 mL of water is(a) P1 > P2 > P3 (b) P3 > P1 > P2

(c) P2 > P1 > P3 (d) P2 > P3 > P1

226. Pressure at which reverse osmosis starts is(a) very high pressure (b) atmospheric pressure(c) osmotic pressure (d) vapour pressure.

227. Osmotic pressure of a solution containing 3 g of glucose in 60 g of water at 15°C is (density = 1 g/mL, molecular weight of glucose = 180 g/mol)(a) 0.34 atm (b) 0.65 atm(c) 6.56 atm (d) 5.57 atm

228. A 5% solution of cane sugar (mol. wt. = 342) is isotonic with 1% solution of a substance X. The molecular weight of X is(a) 34.2 (b) 171.2(c) 68.4 (d) 136.8

229. The hard shell of an egg is dissolved in acetic acid, and then egg is placed in saturated solution of NaCl. Choose the correct statement.(a) The egg will shrink.(b) The egg will become harder.(c) The egg will swell.(d) There will be no change in the size of egg.

230. Two solutions of KNO3 and CH3COOH are prepared separately. Molarity of both is 0.1 M and osmotic pressures are P1 and P2 respectively. The correct relationship between the osmotic pressures is(a) P2 > P1 (b) P2 = P1

(c) P1 > P2 (d) P

P PP

P P1

1 2

2

1 2−=

+

231. A plant cell shrinks when it is kept in(a) a hypotonic solution(b) a hypertonic solution(c) a solution isotonic with the cell sap(d) water.

232. 20 g of a solute was dissolved in 500 mL of water and osmotic pressure of the solution was found to be 600 mm of Hg at 15 °C. Molecular weight of the solute is(a) 1000 g mol–1 (b) 1200 g mol–1

(c) 1400 g mol–1 (d) 1800 g mol–1

233. Which statement is incorrect for osmotic pressure (p), volume (V) and temperature (T)?

(a) p ∝ 1V

if T is constant

(b) p ∝ V if T is constant(c) p ∝ T if V is constant(d) pV is constant if T is constant

234. The osmotic pressure of a decinormal solution of BaCl2 in water is(a) inversely proportional to its Celsius temperature(b) inversely proportional to its absolute temperature(c) directly proportional to its Celsius temperature(d) directly proportional to its absolute temperature.

235. After swimming for a long time in salt water the skin of one’s finger tips wrinkles. Which one of the following property is responsible for this observation?(a) Osmosis (b) Dialysis(c) Electrodialysis (d) Coagulation


236. The osmotic pressure of which solution is maximum? (consider that deci-molar solution of each is 90% dissociated)(a) Aluminium sulphate(b) Barium chloride(c) Sodium sulphate(d) A mixture of equal volumes of (b) and (c)

237. The osmotic pressure of 0.4% urea solution is 1.66 atm and that of 3.42% solution of sugar is 2.46 atm. When both the solution are mixed, then the osmotic pressure of the resultant solution will be(a) 1.64 atm (b) 2.46 atm(c) 2.06 atm (d) 0.82 atm

238. If a 0.1 M solution of glucose (mol. wt. 180 g mol–1) and 0.1 M solution of urea (mol. wt. 60 g mol–1) are placed on the two sides of a semipermeable membrane to equal heights, then it will be correct to say that(a) there will be no net movement across the

membrane(b) glucose will flow across the membrane into urea

solution(c) urea will flow across the membrane into glucose

solution(d) water will flow from urea solution into glucose

solution.

239. Solutions containing 1.63 g of boric acid in 450 mL and 20 g of sucrose (molecular mass = 342) per litre are isotonic. The molar mass of boric acid is

(a) 342 1 63

20× .

(b) 1 63 1000 342

20 450. × ×

×

(c) 1 63 342 450

1000 20. × ×

× (d)

20 342 4501000 1 63× ×

× .

240. The osmotic pressure of a solution at 276 K is 2.5 atm. Its osmotic pressure at 546 K under similar conditions will be(a) 0.5 atm (b) 1.0 atm(c) 2.5 atm (d) 5.0 atm

241. 0.5 M solution of urea is isotonic with(a) 0.5 M NaCl solution(b) 0.5 M sugar solution(c) 0.5 M BaCl2 solution(d) 0.5 M solution of benzoic acid in benzene.

242. The osmotic pressure of a solution at 0°C is 4 atmospheres. What will be its osmotic pressure at 273°C under similar conditions?(a) 4 atm (b) 2 atm(c) 8 atm (d) 1 atm

243. The molar mass (M2) of W2 g solute and the osmotic pressure (p) of the solution prepared in V litres by the solute at temperature T has the following relationship

(a) M2 = W RT

V2p

(b) M2 = W RT2p

(c) M2 = mRTp

(d) M2RT = p

244. If osmotic pressure of 1 M of the following in water can be measured, which one will show the maximum osmotic pressure?(a) AgNO3 (b) MgCl2(c) (NH4)3PO4 (d) Na2SO4

245. Which of the following has the highest osmotic pressure?(a) M/10 HCl (b) M/10 urea(c) M/10 BaCl2 (d) M/10 glucose

246. According to van’t Hoff — Avogadro’s law, volume occupied by a solution is(a) directly proportional to molar mass of solute(b) inversely proportional to mass of solute(c) directly proportional to number of moles of

solute(d) inversely proportional to number of molecules

of solute.


247. Abnormal colligative properties are observed only when the dissolved non-volatile solute in a given dilute solution(a) is a non-electrolyte(b) offers an intense colour(c) associates or dissociates(d) offers no colour.

248. The expression to compute molar mass of a solute from the elevation of boiling point of a solvent is (where the various symbols have their usual meanings)

(a) MKT

WW

b

b2

1

2=D

(b) MT

KWW

b

b2

2

1=D

(c) MKT

WW

b

b2

2

1=D

(d) MT

KWW

b

b2

1

2=D

249. Abnormal molar mass is produced by(a) association of solute(b) dissociation of solute(c) both association and dissociation of solute(d) separation by semipermeable membrane.


250. How many grams of KCl should be added to 1000 g of water, so that the freezing point reduces to –10°C? (Kf for water = 1.86 °C kg mol–1 molar mass of KCl= 74.5 g/mole).(a) 74.5 g (b) 745 g(c) 268 g (d) 199.66 g

251. Acetic acid dissolved in benzene shows a molecular mass of(a) 30 (b) 60(c) 120 (d) 180

252. 2C6H5COOH → (C6H5COOH)2 and 2CH3COOH → (CH3COOH)2 represents

(a) association (b) polymerisation(c) condensation (d) evaporation.

253. The molecular mass of acetic acid dissolved in water is 60 and when dissolved in benzene it is 120. This difference in behaviour of CH3COOH is because(a) water prevents association of acetic acid(b) acetic acid does not dissolve fully in water(c) acetic acid fully dissolves in benzene(d) acetic acid does not ionize in benzene.

2.12 van’t Hoff Factor

254. The van’t Hoff factor i for a dilute aqueous solution of sucrose is(a) zero (b) 1.0(c) 1.5 (d) 2.0

255. If pobs = observed colligative property andpcal = theoretical colligative property assuming normal behaviour of solute then van’t Hoff factor (i) is given by(a) i = pobs × pcal (b) i = pobs + pcal

(c) i = pobs – pcal (d) i = pp

obs

cal

256. If a is the degree of dissociation of Na2SO4 the van’t Hoff factor (i) used for calculating molecular mass is(a) 1 + a (b) 1 – a(c) 1 + 2a (d) 1 – 2a

257. 0.2 m aqueous solution of KCl freezes at –0.680°C. The van’t Hoff factor and observed osmotic pressure of solution at 0°C (Kf = 1.86 K kg mol–1) are respectively(a) 1.83, 8.19 atm (b) 8.13, 9.18 atm(c) 3.18, 19.8 atm (d) 3.81, 89.1 atm

258. 0.01 molal aqueous solution of K3[Fe(CN)6] freezes at –0.062°C. What is the percentage dissociation of solute, Kf for water is 1.86 K kg mol–1?

(a) 87% (b) 78%(c) 89% (d) 98%

259. 8 g of benzoic acid when dissolved in 100 g of benzene lowers its freezing point by 1.62 K. What is the degree of association of benzoic acid if it, forms dimers in benzene, Kf for Benzene is 4.9 K kg mol–1 (C = 12, H = 1, O = 16)?(a) 99.16% (b) 9.9%(c) 91.6% (d) 9.16%

260. 0.6 mL of glacial acetic acid with density 1.06 g mL–1 is dissolved in 1 kg water and the solution froze at –0.0205°C. What is the van’t Hoff factor if Kf for water is 1.86 K kg mol–1?(a) 10.41 (b) 2.41(c) 1.041 (d) 2

261. The van’t Hoff factor i for a 0.2 molal aqueous solution of urea is(a) 0.2 (b) 0.1(c) 1.2 (d) 1.0

262. van’t Hoff factor is(a) less than one in case of dissociation(b) more than one in case of association(c) always less than one(d) less than one in case of association.

263. Which of the following compounds has van’t Hoff factor ‘i’ equal to 2 for dilute solution?(a) K2SO4 (b) NaHSO4(c) Sugar (d) MgSO4

264. van’t Hoff factor (i) is the ratio of(a) observed molar mass to theoretical molar mass(b) observed value of colligative property to

theoretical value(c) theoretical value of colligative property to

observed value(d) none of these.

265. The van’t Hoff factor will be highest for(a) sodium chloride (b) magnesium chloride(c) sodium phosphate (d) urea.

266. The depression in freezing point for 1 M urea, 1 M glucose and 1 M NaCl are in the ratio(a) 1 : 1 : 2 (b) 3 : 2 : 2(c) 2 : 1 : 1 (d) 1 : 1 : 1

267. The degree of dissociation (a) of a weak electrolyte AxBy is related to van’t Hoff factor (i) by the expression

(a) a = −+ −i

x y1

1( ) (b) a = −

+ +i

x y1

1

(c) a =+ −−

x yi

11

(d) a = + +−

x yi

11


268. The van’t Hoff factor i for 0.2 m aqueous glucose solution is(a) 0.2 (b) 0.4(c) 0.6 (d) 1.0

269. The van’t Hoff factor calculated from association data is always ____ than calculated from dissociation data.(a) less (b) more(c) same (d) more or less

270. If van’t Hoff factor for dissolution of Ca(NO3)2 in water is 2.5, then degree of dissociation is(a) 25% (b) 50%(c) 75% (d) 84%

271. The substance A when dissolved in solvent B gave molar mass corresponding to A3. The van’t Hoff factor will be

(a) 12

(b) 1

(c) 2 (d) 13

272. Which of the following compounds corresponds to van’t Hoff factor (i) to be equal to 2 for dilute solution?(a) Na2SO4 (b) AlCl3(c) Glucose (d) BaSO4

273. Which one of the following salt, will have the same value of van’t Hoff factor (i) as that of K4[Fe(CN)6]?(a) Fe2(SO4)3 (b) NaNO3

(c) Ca(NO3)2 (d) K2SO4

274. The van’t Hoff factor 0.1 M CaCl2 solution is 2.74. The degree of dissociation is(a) 61% (b) 87%(c) 100% (d) 54%

275. What will be the ratio of any of the colligative properties of 1.0 m aqueous solutions of NaCl, Na2SO4 and K4[Fe(CN)6] [Assume that solute completely (100%) dissociates in the solution](a) 2 : 3 : 4 (b) 1 : 2 : 4(c) 2 : 3 : 5 (d) 1 : 3 : 5

LEVEL - 2

1. During osmosis, flow of water through a semipermeable membrane is(a) from solution having lower concentration only(b) from solution having higher concentration

only(c) from both sides of semipermeable membrane

with equal flow rates(d) from both sides of semipermeable membrane

with unequal flow rates.

2. Find the percentage of aqueous cane sugar solution which will have the same freezing point as that of 3% aqueous solution of urea (molecular weight of urea = 60 g mol–1, and molecular weight of cane sugar = 342 g mol–1).(a) 17.1% (b) 16.6%(c) 77% (d) 20%

3. Which of the following salts has the same value of vant’ Hoff factor as that of K3[Fe(CN)6]?(a) Na2SO4 (b) Al(NO3)3

(c) Al2(SO4)3 (d) Fe3O4

4. If equimolar solutions of CaCl2 and AlCl3 in water have boiling point of T1 and T2 respectively then(a) T1 > T2 (b) T2 > T1

(c) T1 = T2 (d) can't say

5. If the osmotic pressure of a dilute solution is 7 × 105 Pa at 273 K then its osmotic pressure at 283 K will be(a) 7.256 × 105 Nm–2 (b) 7.562 × 104 Nm–2

(c) 7.256 × 103 Nm–2 (d) 7.652 × 105 Nm–2

6. The solubility of a gas in liquid increases with(a) increase in temperature(b) reduction of gas pressure(c) decrease in temperature(d) amount of liquid taken.

7. Arrange the following aqueous solutions in the order of their increasing boiling points(i) 10–4 M NaCl (ii) 10–4 M Urea (iii) 10–3 M MgCl2 (iv) 10–2 M NaCl(a) (i) < (ii) < (iv) < (iii)(b) (ii) < (i) = (iii) < (iv)(c) (ii) < (i) < (iii) < (iv)(d) (iv) < (iii) < (i) = (ii).

8. A solution containing 0.5126 g naphthalene (mol. mass = 128) in 50 g of carbon tetrachloride yields a boiling point elevation 0.402°C while a solution of 0.6216 g of an unknown solute in the same weight of sample solvent gives a boiling point elevation of 0.647°C. The molecular mass of unknown solute is(a) 98.48 g mol–1 (b) 46.25 g mol–1

(c) 96.46 g mol–1 (d) 46.52 g mol–1


9. If 0.5 m solution of Ca(NO3)2 and 0.75 m solution of KOH is taken, then the depression in freezing point is(a) greater in Ca(NO3)2 because number of ions are

greater(b) greater in KOH because concentration is high(c) equal in both and freezing point is less than 0°C

because ionic concentration is same(d) equal to 0°C in both because ionic concentration

is negligible.

10. CrCl3·6NH3 can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558°C. Assuming 100% ionisation of the complex and co-ordination number of Cr as six, the complex will be (Kf for water = 1.86 K kg mol–1)(a) [Cr(NH3)6]Cl3 (b) [Cr(NH3)5Cl]Cl2(c) [Cr(NH3)4Cl2]Cl (d) [Cr(NH3)3Cl3]Cl

11. A 0.0020 m aqueous solution of an ionic compound [Co(NH3)5(NO2)]Cl freezes at –0.00732°C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = 1.86°C/m)(a) 3 (b) 4(c) 1 (d) 2

12. Pure benzene freezes at 5.3°C. A solution of 0.223 g of phenylacetic acid (C6H5CH2COOH) in 4.4 g of benzene (Kf = 5.12 K kg mol–1) freezes at 4.47°C. From this observation, one can conclude that(a) phenylacetic acid exists as such in benzene(b) phenylacetic acid undergoes partial ionisation

in benzene(c) phenylacetic acid undergoes complete ionisation

in benzene(d) phenylacetic acid dimerises in benzene.

13. What is the molarity of H2SO4 solution, that has a density 1.84 g/cc at 35°C and contains 98% H2SO4 by weight?(a) 18.4 M (b) 18 M(c) 4.18 M (d) 8.14 M

14. Which one of the following statements is false?(a) The correct order of osmotic pressure of

0.01 M aqueous solution of each compound is BaCl2 > KCl > CH3COOH > sucrose.

(b) The osmotic pressure (p) of a solution is given by the equation, p = MRT, where M is the molarity of the solution.

(c) Raoult’s law states that the vapour pressure of a component over a solution is proportional to its mole fraction.

(d) Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression.

15. When 2 g of a non-volatile solute was dissolved in 90 g of benzene the boiling point of benzene is raised by 0.88 K. Which of the following may be the solute? (Kb for benzene = 2.53 K kg mol–1)(a) CO(NH2)2 (b) C6H12O6

(c) NaCl (d) None of these.

16. A solution containing 25.6 g of sulphur dissolved in 1000 g of naphthalene whose melting point is 80.1°C gave a freezing point lowering of 0.68°C. Calculate formula of sulphur (Kf for naphthalene = 6.8 K m–1).(a) S6 (b) S4

(c) S8 (d) S2

17. Phenol associates in benzene to a certain extent to form dimer. A solution containing 2.0 × 10–2 kg of phenol in 1.0 kg of benzene has its freezing point decreased by 0.69 K. The degree of association of phenol is (Kf for benzene = 5.12 K kg mol–1)(a) 73.4 (b) 50.1(c) 42.3 (d) 25.1

18. The order of increasing freezing point of C2H5OH, Ba3(PO4)2, Na2SO4, KCl and Li3PO4 is(a) Ba3(PO4)2 < Na2SO4 < Li3PO4 < C2H5OH < KCl(b) Ba3(PO4)2 < C2H5OH < Li3PO4 < Na2SO4 < KCl(c) C2H5OH < KCl < Na2SO4 < Ba3(PO4)2 < Li3PO4

(d) Ba3(PO4)2 < Li3PO4 < Na2SO4 < KCl < C2H5OH

19. Calculate the weight of ethylene glycol (an effective antifreeze) that must be added to 25 litre water to protect its freezing at –24°C. (Kf = 1.86 °C m–1)(a) 20 kg (b) 322.5 kg(c) 200 kg (d) 32.25 kg

20. 19.5 g of monofluoroacetic acid was dissolved in 0.5 kg of water. The freezing point of solution was observed to be –1.0°C. The van’t Hoff factor, degree of dissociation and dissociation constant of acid are respectively (Atomic masses C = 12, H = 1, F = 19, O = 16 and Kf = 1.86 K kg mol–1)(a) 1.076, 0.076 and 0.003126(b) 0.076, 1.076 and 0.003126(c) 0.003126, 0.076 and 1.076(d) None of these.


21. 10 g of monochlorobutyric acid was dissolved in 250 g of water. If dissociation constant of the acid is 1.45 × 10–3 and Kf = 1.86 K kg mol–1, the depression of the freezing point will be (C = 12, Cl = 35, H = 1, O = 16)(a) 0.56 K (b) 0.65 K(c) 0.065 K (d) 0.056 K

22. Which of the following solutions is a 1 M solution? (C = 12, H = 1, O = 16, Ca = 39.98, Cl = 35.5,Na = 23)(a) 0.46 g of C2H5OH in 100 mL of solution(b) 110.98 g of CaCl2 in 1000 mL of solution(c) 0.23 g of CH3OH in 100 mL of solution(d) 5.85 g of NaCl in 1000 mL of solution

23. A solution contains 25% H2O, 25% C2H5OH and 50% CH3COOH by mass. The mole fraction of H2O would be(a) 0.25 (b) 2.5(c) 0.502 (d) 5.03

24. Match the terms given in Column I with the type of solutions given in Column II.

Column I Column IIA. Soda water 1. A solution of gas in solidB. Sugar solution 2. A solution of gas in gasC. German silver 3. A solution of solid in liquidD. Air 4. A solution of solid in solidE. Hydrogen gas

in palladium5. A solution of gas in liquid

6. A solution of liquid in solid A B C D E(a) 5 6 4 2 1(b) 5 3 4 2 1(c) 1 2 3 4 5 (d) 1 2 3 4 6

25. Which statement is true for solution of 0.020 M H2SO4?(a) 2 litre of the solution contains 0.20 mole of SO4

2–

(b) 2 litre of the solution contains 0.080 mole of H3O+

(c) 1 litre of the solution contains 0.020 mole of H3O+

(d) 1 litre of the solution contains 0.04 mole of SO42–

26. The mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80% is(a) 20 g (b) 30 g(c) 10 g (d) 15 g

27. The partial pressure of ethane over a solution containing 6.56 × 10–2 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas?

(a) 0.762 bar (b) 76.2 bar(c) 0.076 bar (d) 7.6 bar

28. How many grams of sulphuric acid is to be dissolved to prepare 200 mL aqueous solution having concentration of [H3O+] ions 1 M at 25°C temperature?[H = 1, O = 16, S = 32 g mol–1](a) 4.9 g (b) 19.6 g(c) 9.8 g (d) 0.98 g

29. Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.(a) 1.854 kg (b) 1.854 g(c) 18.54 kg (d) 18.54 g

30. How much mL of a 0.1 M HCl is required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of the two?(a) 15.78 mL (b) 157.8 L(c) 157.8 mL (d) 15.78 L

31. 20 g of a binary electrolyte (mol. wt. = 100) are dissolved in 500 g of water. The depression in freezing point of the solution is 0.74°C (Kf = 1.86 K molality–1). The degree of the ionisation of the electrolyte is(a) 0% (b) 100%(c) 75% (d) 50%

32. Which of the following aqueous solutions produce the same osmotic pressure?(i) 0.1 M NaCl solution(ii) 0.1 M glucose solution(iii) 0.6 g urea in 100 mL solution(iv) 1.0 g of a non-electrolyte solute (X) in 50 mL solution (molar mass of X = 200)(a) (i), (ii), (iii) (b) (ii), (iii), (iv)(c) (i), (ii), (iv) (d) (i), (iii), (iv)

33. A solution containing 30 g of a non-volatile solute exactly in 90 g water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is then added to solution, the new vapour pressure becomes 2.9 kPa. The molecular mass of the solute and the vapour pressure of water at 298 K are respectively(a) 25 g and 253 kPa (b) 2.5 g and 2.53 kPa(c) 23 g and 3.53 kPa (d) 2.3 g and 355 kPa

34. Two elements A and B form compounds having molecular formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is


5.1 K kg mol–1. The atomic masses of A and B are respectively(a) 35.29 u and 62.24 u (b) 24.84 u and 43.03 u(c) 21.29 u and 39.49 u (d) 20 u and 40 u

35. The depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water is (Ka = 1.4 × 10–3, Kf = 1.86 K kg mol–1)(a) 0.25°C (b) 0.35°C(c) 0.55°C (d) 0.65°C

36. 1 mole each of the following solutes are taken in 5 moles of water.A. NaCl B. K2SO4

C. Na3PO4 D. glucoseAssuming 100% ionisation of the electrolyte, relative decrease in vapour pressure will be in the order(a) A < B < C < D (b) D < C < B < A(c) D < A < B < C (d) Equal

37. How much C2H5OH must be added to 1.0 L of H2O, so that solution should not freeze at –4°F?[Kf (C2H5OH) = 1.86°C/m](a) < 10.75 g (b) > 494.5 g(c) < 20 g (d) 494.5 g

38. Assume that 0.1 molal solutions of sodium chloride, barium chloride, sodium phosphate and aluminium sulphate are all 100% dissociated. The solution having the highest boiling point is that of(a) NaCl (b) BaCl2(c) Na3PO4 (d) Al2(SO4)3

39. In a 0.2 molal aqueous solution of a weak acid HX, the degree of ionisation is 0.3. Taking Kf for water as 1.85 K kg mol–1, the freezing point of the solution will be nearest to(a) –0.481°C (b) –0.360°C(c) –0.260°C (d) +0.480°C

40. 58.5 g of NaCl and 180 g of glucose were separately dissolved in 1000 mL of water. Identify the correct statement regarding the elevation of boiling point of the resulting solutions.(a) NaCl solution will show higher elevation of

boiling point.(b) Glucose solution will show higher elevation of

boiling point.(c) Both the solutions will show equal elevation of

boiling point.(d) The boiling point elevation will be shown by

neither of the solutions.

41. Which one of the following pairs of solutions are isotonic?(a) 0.1 M urea and 0.1 M NaCl(b) 0.1 M urea and 0.2 M MgCl2(c) 0.1 M NaCl and 0.1 M Na2SO4

(d) 0.1 M Ca(NO3)2 and 0.1 M Na2SO4

42. Which of the following statements is false?(a) The value of molal depression constant depends

on the nature of solvent.(b) Relative lowering of vapour pressure is a

dimensionless quantity.(c) Relative lowering of vapour pressure is equal to

mole fraction of solvent.(d) Equimolal solutions of different non-electrolyte

solutes would elevate the boiling point to the same extent.

43. The freezing point of equimolal aqueous solution will be highest for _____(a) C6H5NH3

+Cl– (b) Ca(NO3)2

(c) La(NO3)3 (d) C6H12O6

44. Repeated measurements of boiling points of separate solutions of 2.36 g of mercurous chloride in 100 g of water produced DTb values in the range of 0.024 to 0.026 K. The atomic weights of mercury and chlorine are 200 and 35.5 respectively. Kb = 0.5 K per mol for water. These data suggest that mercurous chloride functions as a/an(a) covalent compound in aqueous medium(b) ionic compound in aqueous medium(c) reducing agent in aqueous medium(d) oxidising agent in aqueous medium.

45. Solutions A, B, C and D are respectively 0.1 M glucose, 0.05 M NaCl, 0.05 M BaCl2 and 0.1 M AlCl3. Which one of the following pairs is isotonic?(a) A and B (b) B and C(c) A and D (d) A and C

46. A solution containing 0.52 g of KCl in 100 g of water froze at –0.25°C. Calculate the percentage ionisation of the salt. (Kf = 1.86 K m–1, At. masses : K = 39.1, Cl = 35.5)(a) 93 (b) 83(c) 73 (d) 63

47. The van't Hoff factor for 0.1 M Ba(NO3)2 solution is 2.74. The degree of dissociation of the salt at this concentration is (a) 91.3% (b) 87%(c) 100% (d) 74%


48. The vapour pressures of two liquids ‘P’ and ‘Q’ are 80 and 60 torr respectively. The total vapour pressure of the solution obtained by mixing 3 moles of P and 2 moles of Q would be(a) 68 torr (b) 140 torr(c) 72 torr (d) 20 torr

49. Ebullioscopic constant of water is 0.52 K kg mol–1. The incorrect statement is(a) 30 g of urea in half a kg water produces 0.26 K

elevation in boiling point(b) 60 g of urea in one kg water produces 0.52 K

elevation in boiling point(c) 90 g of glucose in half a kg water produces

0.52 K elevation in boiling point(d) 120 g of urea in one kg water produces 1.04 K

elevation in boiling point.

50. A sugar solution in water is expected to freeze(a) at 273 K (b) below 273 K(c) above 273 K (d) at 298 K.

51. An aqueous solution of a weak monobasic acid containing 0.1 g in 21.7 g of water freezes at 272.813 K. If the value of Kf for water is 1.86 K/m, what is the molecular mass of the monobasic acid?(a) 50.0 g/mol (b) 46.2 g/mol(c) 55.5 g/mol (d) 25.4 g/mol

52. Which one of the following solutions are isotonic?(a) 3.42 g of sugar in 1000 cm3 of water and 0.18 g

of glucose in 1000 cm3 of water(b) 3.42 g of sugar in 1000 cm3 of water and 0.18 g

of glucose in 100 cm3 of water(c) 3.42 g of sugar in 1000 cm3 of water and 0.585 g

of NaCl in 1000 cm3 of water(d) 3.42 g of sugar in 1000 cm3 of water and 5.85 g

of NaCl in 100 cm3 of water

53. Two solutions X and Y are separated by a semipermeable membrane. If the solvent flows from X to Y, then(a) X is more concentrated than Y(b) X is less concentrated than Y(c) both X and Y have the same concentration(d) both X and Y get diluted.

54. Which of the following statements is not correct?(a) Osmotic pressure is directly proportional to

molar concentration.(b) Hypertonic solutions have lower concentrations

with respect to reference solution.(c) Isotonic solutions have same molar concentration.(d) Osmotic pressure depends upon temperature.

55. When 5% solution of sucrose (molar mass = 342) is isotonic with 1% solution of a compound ‘A’, the molar mass of A is(a) 180 g (b) 68.4 g(c) 18 g (d) 32 g

56. A solution is obtained by mixing 300 g of a 25% solution and 400 g of a 40% solution by mass. Calculate the mass percentage of solute in the resulting solution.(a) 33.57 (b) 66.43(c) 87.23 (d) 19.24

57. Given Kb and Kf of water are 0.52 and 1.86 K m–1. An aqueous solution freezes at –0.186°C. What is the boiling point of the solution?(a) 0.52°C (b) 100.52°C(c) 0.052°C (d) 100.052°C

58. Pure water as well as separate equimolal and dilute aqueous solutions of NaCl, K3[Fe(CN)6] and K4[Fe(CN)6] represented respectively as I, II, III and IV are available. Which of the following statements is correct, assuming 100% dissociation of all solutes?(a) I < II < III < IV is the increasing order of freezing

points(b) IV < III < II < I is the increasing order of freezing

points(c) IV < III < II < I is the increasing order of boiling

points(d) IV < III < II is the increasing order of van’t Hoff

factors.

59. Solution A contains 1 g of urea in 100 mL of water and solution B contains 2 g of glucose in 100 mL of water. Now,(a) boiling point of solution A will be less than

solution B(b) freezing point of solution A will be lower than

solution B(c) both will have same freezing point and boiling

point(d) osmotic pressure of solution A will be less than

solution B.

60. Which of the following statements is correct?(a) Lowering of vapour pressure takes place only in

ideal solutions(b) Lowering of vapour pressure does not depend

upon the solvent at a given concentration of the solute.

(c) Lowering of vapour pressure depends upon the nature of solute

(d) Relative lowering of vapour pressure does not depend upon the solvent at a given concentration of the solute.


2015

1. If M, W and V represent molar mass of solute, mass of solute and volume of solution in litres respectively, which among the following equations is true?

(a) p =MWR

TV (b) p =

TMRWV

(c) p =TWRVM

(d) p =TRVWM

(MH-CET)

2. Molarity is defined as(a) the number of moles of solute dissolved in one dm3

of the solution(b) the number of moles of solute dissolved in 1 kg of

solvent(c) the number of moles of solute dissolved in 1 dm3 of

the solvent(d) the number of moles of solute dissolved in

100 mL of the solvent. (MH-CET)

3. van’t Hoff factor of centimolal solution of K3[Fe(CN)6] is 3.333. Calculate the percent dissociation of K3[Fe(CN)6].(a) 33.33 (b) 0.78(c) 78 (d) 23.33 (MH-CET)

4. What is the mole fraction of the solute in a 1.00 m aqueous solution?(a) 1.770 (b) 0.0354(c) 0.0177 (d) 0.177 (AIPMT)

5. The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol–1) of the substance is (a) 128 (b) 488(c) 32 (d) 64 (JEE Main)

2014

6. What is the molality of a solution containing 200 mg of urea (molar mass = 60 g mol–1) dissolved in 40 g of water ?(a) 0.0825 (b) 0.825(c) 0.498 (d) 0.0013 (MH-CET)

7. Solubility of which among the following substances in water increases slightly with rise in temperature?(a) Potassium bromide (b) Potassium chloride(c) Potassium nitrate (d) Sodium nitrate

(MH-CET)

8. Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and 0.125 M Na3PO4(aq) at 25 °C. Which statement is true about these solutions, assuming all salts to be strong electrolytes?(a) 0.500 M C2H5OH(aq) has the highest osmotic

pressure.(b) They all have the same osmotic pressure.(c) 0.100 M Mg3(PO4)2(aq) has the highest osmotic

pressure.(d) 0.125 M Na3PO4(aq) has the highest osmotic

pressure. (JEE Main)

2012

9. pA and pB are the vapour pressure of pure liquid components, A and B, respectively of an ideal binary solution. If xA represents the mole fraction of component A, the total pressure of the solution will be(a) pA + xA(pB – pA) (b) pA + xA(pA – pB)(c) pB + xA(pB – pA) (d) pB + xA(pA – pB) (AIPMT)

10. Kf for water is 1.86 K kg mol–1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8°C?(a) 93 g (b) 39 g(c) 27 g (d) 72 g (AIEEE)

11. The density of a solution prepared by dissolving 120 g of urea (mol. mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity of this solution is(a) 1.78 M (b) 1.02 M(c) 2.05 M (d) 0.50 M (AIEEE)

CompetitiveExams


2011

12. An aqueous solution of urea containing 18 g urea in 1500 cm3 of the solution has a density equal to 1.052. If the molecular weight of urea is 60, the molarity of the solution is(a) 0.200 (b) 0.192(c) 0.100 (d) 1.200 (MH-CET)

13. 34.2 g of cane sugar is dissolved in 180 g of water. The relative lowering of vapour pressure will be(a) 0.0099 (b) 1.1597(c) 0.840 (d) 0.9901 (MH-CET)

14. Mole fraction of the solute in a 1.00 molal aqueous solution is(a) 0.1770 (b) 0.0177(c) 0.0344 (d) 1.7700 (AIPMT)

15. A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be(a) – 0.18°C (b) – 0.54°C(c) – 0.36°C (d) – 0.24°C

(AIPMT Mains)

16. 200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57 × 10–3 bar. The molar mass of protein will be(R = 0.083 L bar mol–1 K–1)(a) 51022 g mol–1 (b) 122044 g mol–1

(c) 31011 g mol–1 (d) 61038 g mol–1

(AIPMT Mains)

17. The freezing point depression constant for water is 1.86°C m–1. If 5.00 g Na2SO4 is dissolved in

45.0 g H2O, the freezing point is changed by 3.82°C. Calculate the van’t Hoff factor for Na2SO4.(a) 2.05 (b) 2.63(c) 3.11 (d) 0.381 (AIPMT)

18. The van’t Hoff factor i for a compound which undergoes dissociation in one solvent and association in other solvent is respectively(a) less than one and greater than one(b) less than one and less than one(c) greater than one and less than one(d) greater than one and greater than one. (AIPMT)

19. The degree of dissociation (a) of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression

(a) a = −+ −i

x y1

1( ) (b) a = −

+ +i

x y1

1( )

(c) a = + −−

( )x yi

11

(d) a = + +−

( )x yi

11

(AIEEE)

20. Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at – 6°C will be : (Kf for water = 1.86 K kg mol–1, and molar mass of ethylene glycol = 62 g mol–1)(a) 804.32 g (b) 204.30 g(c) 400.00 g (d) 304.60 g (AIEEE)

21. A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of methyl alcohol in the solution?(a) 0.100 (b) 0.190(c) 0.086 (d) 0.050

(AIEEE)


LEVEL - 11. (a) 2. (c) 3. (a)

4. (d) 5. (a) 6. (a)

7. (c) 8. (d) 9. (a)

10. (c) : Smoke is an aerosol, with gas (air) as the dispersion medium and solid (soot particles mostly) as the dispersed phase.

11. (c) 12. (b)

13. (a) : Mixture Solute Solvent

(a) Chloroform in N2 gas liquid gasCO2 in water gas liquid

(b) Brass solid solidCamphor in N2 gas solid gas

(c) Sodium amalgam liquid solidMoist air liquid gas

(d) H2 gas in Pd metal gas solidMixture of N2 and O2 gas gas

14. (b) 15. (c) 16. (a)

17. (d) : Precentage by volume of C2H5OH (v/V)

= ×Volume of solute

Volume of solution 100

= × =22

220100 10%

18. (d) : M1V1 = M2V2

2 × 0.1 = 110

× V2 ⇒ V2 = 2 dm3

19. (a) : Molality Moles of solute1000 g of solvent

=

20. (a) : M = WM V

2

23

1000×× ( )incm

= 10 100040 250××

= 1 M

21. (a) : m = nW2

1

1000 0 5400

×=

in g.

× 1000 = 1.25 m

22. (a)

23. (b) : Molarity =××

WM V2

2

1000

Molarity M= ××

=36 1000180 500

0 4.

(Qdensity of water = 1 g/cm3)

Hints & Explanations24. (b) : Molality =

Number of moles of soluteWeight of solvent (in g)

××1000

m = × =64818

1000648

55 5. m

25. (b) : Moles of C2H5OH = 2546

0 54= .

Moles of CH3COOH = 5060

0 83= .

% of H2O = 100 – (50 + 25) = 25

\ Moles of H2O = 2518

1 39= .

Total number of moles = 2.76

Mole fraction of acetic acid = 0 832 76

0 301..

.=

26. (b) : ppm = 0 5 1010

6

7. × = 0.05

27. (b) : Percentage by mass of urea (w/W)

= ×mass of ureamass of urea + mass of water

100

=+

× = =6

6 500100 600

5061 186

gg g

by mass. %

28. (c) : Percentage by volume of ethyl alcohol

= ×volume of ethyl alcohol

volume of solution100

= ×58

454100

3

3cm

cm = 12.78% by volume

29. (d) : Moles of ethyl alcohol (n2)

= mass of ethyl alcohol

molar mass of ethyl alcohol

= =−

23

460 51

g

g molmol.

Moles of water (n1) =mass of water

molar mass of water

= −54

18 1g

g mol= 3.0 mol

Mole fraction of ethanol = =+

=+

x nn n2

2

1 2

0 53 0 0 5

.. .

= 0.1429

Mole fraction of water = =+

=+

xn

n n11

1 2

3 03 0 0 5

.. .

= 0.8571and x1 + x2 = 0.8571+ 0.1429 = 1


30. (c) : 35% (w/W) solution of ethylene glycol in water means it has 35 g of ethylene glycol and 65 g of water.Number of moles of ethylene glycol

= Mass of ethylene glycolMolar mass of ethylene glycol

= 3562

= 0.56 moles

Number of moles of water =Mass of water

Molar mass of water

= 6518

= 3.61 moles

\ Mole fraction of ethylene glycol

=

+= =0 56

0 56 3 610 564 17

0 1343.. .

.

..

31. (c) : Mass of NaOH = 1.6 × 10–3 kg,Volume of solution = 500 cm3 = 0.5 dm3,Molar mass of NaOH = 40 × 10–3 kg mol–1

Molarity of NaOH solution

=×−

mass of NaOH in kg

molar mass of NaOH in kg mol volume o1 ff

solution in dm3

=

×× ×

=−

− −−1 6 10

40 10 0 50 08

3

3 1 33.

..

kg

kg mol dmmol dm

32. (b) : Molar mass of urea, NH2CONH2 = 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol–1

Mass of solvent = 100 g = 0.1 kg

Moles of urea = =−11 11

600 18521

..

g

g molmol

Molality of solution = moles of ureamass of solvent in kg

= = −0 1852

0 11 852 1.

..

molkg

mol kg

33. (c) : Molar mass of sugar, C12H22O11 = (12 × 12) + (1 × 22) + (11 × 16) = 342 g mol–1 = 342 × 10–3 kg mol–1

Mass of water in syrup = mass of syrup – mass of sugar = 214.2 g – 34.2 g = 180 g = 180 × 10–3 kg

Number of moles of sugar = mass of sugarmolar mass of sugar

= 34 2 10

342 10

3

3 1. ×

×

−

− −kg

kg mol= 0.1 mol

Moles of water = mass of water

molar mass of water

= 180 10

18 10

3

3 1×

×

−

− −kg

kg mol = 10 moles

Molality = moles of sugarmass of solvent in kg

= 0 1

180 10 3. mol

kg× − = 0.556 mol kg–1

Mole fraction of sugar in syrup

= moles of sugar

moles of sugar + moles of water

= 0 1

0 1 100 1

10 10 0099.

...

.+

= =

34. (a) : H2SO4 solution is 27% by massHence, if mass of H2SO4 is 27 gThen mass of H2O is (100 – 27) = 73 gMolality of H2SO4

= mass of H SO

molar mass of H SO mass of solvent in kg2 4

2 4 ×

= 27 10

98 10 73 10

3

3 1 3×

× × ×

−

− − −kg

kg mol kg = 3.77 mol kg–1

Density of solution = Mass of solute

Volume of solution

Therefore, volume of solution = 100

1 198 3

g

g cm. −

= 83.47 cm3 = 83.47 × 10–3 dm3

Molarity of solution

= Mass of solute

Molar mass of solute Volume of solution in d× mm3

= 27 10

98 10 83 47 10

3

3 1 3 3×

× × ×

−

− − −kg

kg mol dm.

= 3.301 mol dm–3

35. (c) : HCl solution is 38% by mass.Hence, 100 g HCl solution contains 38 g HCl and (100 – 38) = 62 g water.Volume of 100 g HCl solution

= Mass of solution

Density of solution = 100

1 1 3g

g cm. −

= 90.91 cm3 = 90.91 × 10–3 dm3


= Mass of HCl

Molar mass of HCl Volume of solution in dm× 3

= 38 10

36 5 10 90 91 10

3

3 1 3 3×

× × ×

−

− − −kg

kg mol dm. .= 11.45 mol dm–3

Moles of HCl , n2 = Mass of HCl

Molar mass of HCl

= 38 1036 5 10

3

3 1×

×

−

− − kg

kg mol. = 1.04 mol


Moles of H2O, n1 = Mass of H OMolar mass of H O

2

2

= 62 10

18 10

3

3 1×

×

−

− −kg

kg mol = 3.44 mol

Mole fraction of HCl, x2 = nn n

2

1 2+=

1 043 44 1 04

.. .+

= 0.232

36. (b) : 15 ppm (by mass) means 15 g chloroform in 106 g of the solution.Mass of solvent ≈ 106 gMolar mass of CHCl3 = 12 + 1 + 3 × 35.5 = 119.5 g mol–1

Number of moles of solute

= Mass of solute

Molar mass of solute= 15

119 5. = 0.126 mol

\ Molality = No. of moles of soluteMass of solvent in g

×1000

= 0 126106. × 1000 = 1.26 × 10–4 m

37. (b) : Mass of glucose = 34.2 gMass of water = 400 gPercentage by mass of glucose in solution (w/W)

= ×Mass of solute

Mass of solution100

=

+× = ×34 2

34 2 400100 34 2

434 2100.

...

= 7.87 % by mass

38. (c) : Mass of water = 500 gPercentage by mass of a solute in solution = 2.38

Percentage by mass of solute = ×Mass of soluteMass of solution

100

2.38 = ×Mass of soluteMass of solution

100

Let mass of solute be ‘x’.

\ 2.38 = x

x +

500

× 100 ⇒ 2.38(x + 500) = 100x

\ 2.38x + 1190 = 100x ⇒1190 = 100x – 2.38 x

\ 1190 = 97.62 x ⇒ x = 119097 62.

= 12.19 g

39. (b) : Density = mass

volumeMass of methyl alcohol = 0.7952 × 4.6 = 3.66 gPercentage by mass of methyl alcohol

= 3 663 66 25 2

.. .+

× 100 = 12.68% by mass

Number of moles of methyl alcohol

= Mass of methyl alcohol

Molar mass of methyl alcohol

= 3 6632. = 0.1144 moles

Number of moles of water

= Mass of water

Molar mass of water= 25 2

18.

= 1.4 moles

Mole fraction of methyl alcohol = 0 1144

0 1144 1 4.

. .+ = 0.0755

40. (c) : Volume of benzene = 12.8 cm3

Volume of xylene = 16.8 cm3

% by volume of solute = Volume of solute

Volume of solution×100

% by volume of benzene = 12 8

12 8 16 8100.

. .+

×

= 43.24% by volume

41. (b) : Solution has 24.8% of HCl, that means 24.8 g of HCl and 75.2 g of H2O are present.Number of moles of HCl

= Mass of HClMolar mass of HCl

moles= =24 836 5

0 6795..

.

Number of moles of H2O

= Mass of H O

Molar mass of H Omoles2

2

75 218

4 1777= =. .

Mole fraction of HCl = no. of moles of HCl

total no. of moles

Mole fraction of HCl = 0 67950 6795 4 1777

0 67954 8572

.. .

.

.+=

= 0.1399

42. (a) : 2 molal aqueous solution is given.It means 2 moles of solute are present in 1000 g of solvent i.e., water.

Number of moles of H2O = MassMolar mass

= 100018

= 55.56 moles

Mole fraction of solute = 2

2 55 560 0347

+=

..

43. (c) : 12.2% HNO3 means 12.2 g of HNO3 and 87.8 g of solvent are present.

No. of moles of HNO3 =Mass of HNO

Molar mass of HNO3

3

12 263

= .

= 0.1937 moles

Molality of HNO3

= Mass of HNO

Molar mass of HNO Mass of solvent in kg3

3 ×

= ×

× × ×=

−

− −12 2 10

63 10 87 8 102 206

3

3 3.

.. mol/kg


Density of solution = Mass of solution

Volume of solution

\ Volume of solution = 1001 038.

= 96.34 cm3

= 96.34 × 10–3 dm3

Molarity

= Mass of solute


= 12 2 1063 10 96 34 10

2 013

3 33.

..×

× × ×=

−

− − mol/dm

44. (b) : 95.8% by mass sulphuric acid is present.

Number of moles of H2SO4 = Mass of H SO

Molar mass of H SO2 4

2 4

= 95 898

0 9776. .= mol

Mass of H2O = 100 – 95.8 = 4.2

Number of moles of H2O = Mass of H O

Molar mass of H O2

2

= 4 218

0 2333. .=

\ Mole fraction of H2SO4 = 0 9776

0 9776 0 2333.

. .+

= 0 97761 2109

0 8073..

.=

Density of solution = Mass of solutionVolume of solution

Volume of solution = Mass of solutionDensity of solution

= 1001 91.

= 52.36 cm3

= 52.36 × 10–3 dm3


= Mass of solute


= 95 898 52 36 10

18 673.

..

× ×=− M

45. (a) : Density of solution = 1.070 g/cm3

Density of solution = Mass of solution

Volume of solution

\ Volume of solution = Mass of solution

Density of solution

= 1001 070.

= 93.46 cm3

\ Volume of solution = 93.46 × 10–3 dm3


= Mass of solute


= 10

40 93 46 102 6753× ×

=−.. M

Molality of solution

= Mass of NaOHMolar mass of NaOH Mass of solvent in kg×

= 10

40 90 102 773

1

× ×=−

−. mol kg

46. (c) : The given aqueous solution of H2SO4 is 4.22 M and has density of 1.21 g/cm3.1.21 g/cm3 = 1.21 g/mLNow, 1000 mL of solution contains 4.22 moles of

H2SO4

\ Mass of 1000 mL of H2SO4 solution = 1000 × 1.21 = 1210 g1 mole of H2SO4 = 98 g of H2SO4

\ 4.22 mole of H2SO4 = 4.22 × 98= 413.56 g of H2SO4

1210 g of H2SO4 solution = 413.56 g of H2SO4

So, 100 g of H2SO4 solution = 34.18 g of H2SO4

\ 100 g of H2SO4 solution contains 34.18 g of H2SO4 and 65.82 g H2O.\ Molality of H2SO4

= Mass of H SO

Molar mass of H SO Mass of solvent in kg2

2

4

4 ×

= ×

× × ×=

−

− −34 18 10

98 10 65 82 105 298

3

3 3.

.. mol/kg

47. (b) 48. (b)

49. (d) : Molality

=

×Mass of solute

Molar mass of solute mass of solvent in kgg

=

× ×=−

2040 500 10

13

mol/kg

50. (a)

51. (d) : Number of moles of H2O = 1

18g/mLg/mol

= 0.0555 mol/mL = 0.0555 × 1000 = 55.5 moles/L

52. (d) : M1 = 0.1 M, M2 = 0.2 MV1 = 100 mL, V2 = 25 mL

Resulting Molarity = M V M V

V V1 1 2 2

1 2

++

= ( . ) ( . ) .0 1 100 0 2 25

100 250 12× + ×

+= M

53. (a) : Molarity = n

VH PO3 4

in LM= × =49

9812

0 25.

54. (d) : Number of moles of Na2SO4 = 7 1142

. = 0.05


Molarity = Number of moles of solute

Volume of solution in L

= 0 050 1

0 5..

.= M

55. (a) : Percentage by volume

= Volume of solute

Volume of solution× =

+

×100 60

60 300100

= 16.66%

56. (d) : 0.8 moles of solute are present in 1000 mL\ 0.1 mole of solute is present in x mL

\ = ⇒ = =0 80 1

1000 10008

125.. x

x mL

57. (c) : Resulting molarity =++

M V M VV V

1 1 2 2

1 2

= × + ×

+= =3 25 4 75

25 75375100

3 75. M

58. (d) : Molecular wt. of Na2CO3 = 106 g/mol1 molar means 1000 mL Na2CO3 solution contains

106 g of Na2CO30.25 M means 1000 mL Na2CO3 solution contains

26.5 g of Na2CO3250 mL of 0.25 M Na2CO3 solution contains

6.625 g of Na2CO3

59. (a) : HCl NaOH M1V1 M2V2 MHCl × 20 = 0.01 × 19.85

MHCl = 0 01 19 8520

0 0099. . .× = M

60. (a) : 58.5 g of NaCl in 1 litre gives 1 molar solution.58.5 g of NaCl in 0.5 litre gives 2 molar solution.\ 5.85 g of NaCl in 0.5 litre gives 0.2 molar solution.

61. (b) : Mmix = M V M V

V V1 1 2 2

1 2

1 2 5 0 5 32 5 3 0

++

= × + ×+

. .

. .

= + = =2 5 1 55 5

45 5

0 73. .. .

. M

62 (a) : Concentration of AgNO3 is 0.03 g/mL. i.e., 1 mL of the solution contains 0.03 g of AgNO3.Thus, to prepare 60 mL of the solution (0.03 × 60)

= 1.8 g of AgNO3 is required.

63. (b) : 0.5 moles of CaCl2 are present in 1000 mL of solution

500 mL of solution contains 0 5

1000500. × moles of CaCl2

= 0.25 moles of CaCl2CaCl2

Ca2+ + 2Cl–

1 mole 2 moles\ 500 mL of solution contains 2 × 0.25 = 0.50 moles of Cl– ions

64. (c) : xA + xB + xC

=+ +

+ +( ) =1 1 0n n n

n n nA B C

A B C .

65. (a)

66. (b) : According to Henry’s law, S = KP⇒ S = 1.1 × 10–3 × 0.20 = 2.2 × 10–4 mol dm–3

67. (a)

68. (d) : According to Henry’s law S = KPK = 1.3 × 10–3 mol dm–3 atm–1

P = 0.22 atmHence, S = 1.3 × 10–3 mol dm–3 atm–1 × 0.22 atm = 2.86 × 10–4 mol dm–3

69. (c) : According to Henry’s law S = KPGiven S = 3.12 × 10–4 mol dm–3, P = 0.24 atm

K = SP= × − −3 12 10

0 24

4 3..

mol dmatm

= 13 × 10–4 mol dm–3 atm–1

= 1.3 × 10–3 mol dm–3 atm–1

70. (c) 71. (d) 72. (c)

73. (b)

74. (c) : Alloy of mercury with other metals is known as amalgam, where mercury is in liquid state.

75. (c) : Stainless steel contains chromium and some nickel.

76. (a) 77. (d) 78. (d)

79. (a)

80. (c) : Relative lowering of vapour pressure is a colligative property, not the vapour pressure.

81. (a) 82. (a) 83. (c)

84. (c) 85. (a) 86. (c)

87. (a)

88. (d) : According to Raoult’s law, the relative lowering in vapour pressure of a dilute solution is equal to mole fraction of the solute present in the solution.

89. (c) 90. (b)

91. (b) : Ptotal = pA + pB = xA p°A + xB p°B

184 35

200 25

= × + × pBo or, 184 120 2

5= + pB

o

⇒ p°B = 160 torr

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