chemistry. solution - i introduction solubility henry’s law different concentration terms ...
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Chemistry
Solution - I
Introduction
Solubility
Henry’s law
Different concentration terms
Vapour pressure
Raoult’s law and its modification
Relative lowering of vapour pressure
Ideal solutions and non-ideal solutions
Maximum and minimum boiling solutions
Session objectives
Solute: Component of solution present in smaller amount.
Solvent: Component of solution present in the larger amount.
Introduction
Solution: a homogenous mixture of two or more substances.
Solubility
For solids Solubility of ionic compounds
in water generally increases with increase in temperature.
For gases The solubility of gases in
water decreases with increase in temperature.
Solubility tends to zero at the boiling point of water.
Maximum amount of solute in grams which can be dissolved in a given amount of solvent (generally 100 g) to form a saturated solution at that particular temperature is known as its solubility
Effect of pressure on solubility of gases
Increase in pressure of the gas above the solutionincreases the solubility of the gas in the solution.
More dilute solution More concentrated solution
Henry’s law
Solubility of a gas in a liquid is proportional to the pressure of the gas over the solution.
C = kP C: molar concentration, P: pressure, k: temperature-dependent constant
Carbonated cold drink is an application of Henry’s law.
moles of solute (mol)Molarity (M)
volume of solution (L)
moles of solute
Molality (m) Weight of solvent (inkg)
Different concentration terms
i
i
nMole fraction(x)
n x1=mole fraction of solvent
x2=mole fraction of solute
Molarity of a solution changes with temperature dueto accompanied change in volume of the solution.
Illustrative Example
Determine the molality of a solution preparedby dissolving 75 g of Ba(NO3)2(s) in 374 g of water at 25oC.
Solution:
Molar mass of Ba(NO3)2 = 261
275
0 287
0 7670 374
3 -1g
Number of moles of Ba(NO )261 g mol
. mole
0.287 moleMolality = . m
. kg
Illustrative Example
Calculate the molality of 1 molar solution of NaOH given density of solution is 1.04 gram/ml.
Solution:
1 molar solution means 1 mole of solute present perlitre of solution.
1
m = ×1000 =1 molal solution.1000
Therefore, mass of 1 litre solution = 1000 x 1.04
= 1040 gram
Mass of solute = 1 x 40 = 40g
Therefore, mass of solvent 1040 – 40 = 1000g
Different concentration terms
Mass of solute% by mass = ×100
Mass of solution
w= %
W
Volume of solute% by volume 100
Volume of solution
v%
V
6mass of soluteParts per million (ppm) = ×10
mass of solution
Illustrative Example
Calculate the concentration of 1 molal solution of NaOH in terms of percentage by mass.
Solution:
1 molal solution means 1 mole (or 40g) NaOH present in 1000g of solvent.
Total mass of solution = 1000 + 40 = 1040g
Therefore, 1040g solution contains 40g NaOH
Therefore, 100g solution contains40
1001040
= 3.84% by mass.
Different Concentration terms
Relation between Molarity (M) and molality (m)
23
mdM
MM1
10
2M Molar mass of solute
d density of solution
Relation between molality(m) and mol-fraction (x2) of solute
2 3
1
1
mx =
10m+
M
Where M = Molar mass of the solvent
12 3
1
mMx
mM 10
Illustrative Example
Calculate the molality and mole fraction of the solute in aqueous solution containing 3.0 g of urea (molecular mass = 60) in 250 g of water.
Mass of solute 1000
MolalityMolecular mass of solute mass of solvent in gram
31000 0.2
60 250
Mole fraction of urea =Moles of urea 3 / 60
0.003593 250Total moles
60 18
Mole fraction of water = 1 – 0.00359 = 0.996
Solution:
Illustrative Example
Calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass.
95% of ethanol by mass means 95 g ethanol present in 100 g of solution.
Hence, mass of water = 100 – 95 = 5 g
Moles of C2H5OH =9546
= 2.07 moles
Moles of water(H2O)=5
= 0.28mol18
Solution:
Mole fraction of C2H5OH = 0.28
= 0.880.28 +2.07
Mole fraction of water = 1 – 0.88 = 0.12
Vapour pressure of solution
Liquid molecules evaporate from the surface
Vapourised molecules condensed to liquid
Both processes reach equilibrium
Po=Pressure exerted by the vapour above the liquid surface at eqm.
vapour pressure of pure
liquid
Factors affecting Vapour PressureNature of liquid:
More volatile liquids exert more pressure on the liquid surface.
Presence of a solute Due to presence of volatile and non-volatile solute, vapour pressure of solution decreases.
Temperature: Increase in temperature increases vapour pressure.
Vapour pressure Temperature
Vapor Pressure of Solution
Some of the solute particles will be near the surface.
Less no. of molecules per unit surface area are involved in equilibrium.
Block solvent molecules from entering the gas phase.
Raoult’s law for non-volatile solute
For highly dilute solutions
po=vapour pressure of pure liquidx1=mol. fraction of solventps=vapour pressure of solution
ps=x1po
Applicable for ideal solution
mix mixH 0 V 0
Here, solute-solute and solvent-solvent interaction exactly equal in magnitude with solute-solvent interaction.
Raoult’s law for non-volatile solute
Relative lowering of vapour pressure
From Raoult’s law,
o ss solvent soluteo
pp x p , 1 x
p
ssolute o
os
o
px 1
p
p p
p
os
soluteo
Relative lowering of v.p,
p px
p
os
o
p p nn Np
n moles of soluteN moles of solvent
os
o
p p nNp
when n 10%
Modification (two volatile liquids)
According to Raoult’s law,for two volatile miscible liquids
o os A A B B
A B
p p x p x
p p (1)
pA Partial vapour pressure of A.
xA Mol fraction of A in liquid phase.
A Bx x 1
Modification (two volatile liquids)
Illustrative Example
Vapour pressure of liquids A and B at a particular temperature are 120 mm and 180 mm of Hg. If 2 moles of A and 3 moles of B are mixed to form an ideal solution, what would be the vapour pressure of the solution?
Solution :
o oA Bp 120 mmHg p 180 mmHg
A Bn 2 mol n 3mol.
A B2 3
x x5 5
o oS A A B Bp p x p x
2 3120 180
5 5
48 108 156
Illustrative Problem
At 40oC, the vapour pressure in torr of methyl alcohol-ethyl alcohol solution is represented by P = 119Xm + 135 where Xm is the mole fraction of methyl alcohol. What are the vapour pressures of pure methyl alcohol & ethyl alcohol ?
Solution
o om m E E
o om m E m
o o om E m E
P = p x +p x
= p x +p 1- x
= p - p x +p
m
oE
o om E
om
Comparing it with
p 119x 135
p 135
p p 119
p 119 135 245 torr
Illustrative Problem
6g of urea is disolved in 90g water at 25oC ? What is vapour pressure of sol. If vapour pressure of water is 40mmHg.
Solution
ps = po x solvent
solventsolvent
total
nX =
n
90/18=
90/18 +6/60
50.980
5 .1
ps = 0.980 x 40 = 39.2 mm Hg
Modification (two volatile liquids)
From Dalton’s law of partial pressure
A A sp = y p - - - (2)
yA=mol. fraction of A in vapour phase
ps=vapour pressure of solution.
From (2)
AA
S
oA A
o oA A B B
py
p
p x
p x p x
o oA A B B
oA A A
oB A
A BoA A
p x p x1y p x
p (1 x )1 x x 1
p x
o oB Bo o
A A A
p p1
p x p
o oB Bo o
A A A A
p p11
y p x p
Modification (two volatile liquids)
Illustrative Problem
An unknown compound is immiscible with water. It is steam distilled at 98.0oCand P = 737 Torr.po
H20 = 707 torr at 98.0oC. This distillate was 75% by weight of water.Calculate the molecular weight of the unknown
Solution
Using Dalton’s law of partial pressurePtotal = 737 torr PoH2O = 707 torr
Pounknown = 737 – 707 = 30 torr.
If water = 100 g the unknown = 75.0 g
22
ounknown unknown
oH OH O
P n 7518
n M 100P
30 75 18707 m 100m 318.15 gmol
Non-ideal solution
Solute-solvent interaction are different than solute-solute and solvent solvent in non ideal solutions.
These do not obey Raoult’s Law.
For non ideal solutions H 0 and V 0
Non ideal solution
For solution showing negative deviation from Raoult's law.
For solution showing positive deviation from Raoult's law.
H 0 and V 0
s idealP >PH 0 and V 0
s idealP P
Azeotropic mixtures
Solution showing positive deviation from Raoult’s form minimum boiling azeotrope
Interaction between A–B < interaction between A–A or B–B
Liquid mixtures which distil without any change in composition are called Azeotropes or Azeotropic mixtures.
Azeotropic mixtures
Solution showing negative deviation from Raoult’s law form maximum boiling azeotropes
Interaction between A – B > interaction between A – A or B – B
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