worked example 11.1 correlating solubility and chemical...

$: WORKED EXAMPLE 11.1 Correlating Solubility and Chemical …programs.clarendoncollege.edu/programs/NatSci/Wiginton... · 2021. 1. 11. · this solution in (a) mole fraction, (b) mass$
Copyright ©2010 by Pearson Education, Inc.General Chemistry: Atoms First

By John E. McMurry and Robert C. Fay

WORKED EXAMPLE 11.1 Correlating Solubility and Chemical

Structure

Strategy

Pentane (C5H12) and 1-butanol (C4H9OH) are organic liquids with similar molecular masses but substantially

different solubility behavior. Which of the two would you expect to be more soluble in water? Explain.

Look at the structures and decide on the kinds of intermolecular forces present

between molecules. The substance with intermolecular forces more like those in

water will be more soluble in water.

Solution

Pentane is a nonpolar molecule and is unlikely to have strong intermolecular

interactions with polar water molecules. 1-Butanol, however, has an –OH part

just as water does and is therefore a polar molecule that can form hydrogen

bonds with water. As a result, 1-butanol is more soluble in water.

PROBLEM 11.1 Arrange the following compounds in order of their expected increasing solubility in water:

Br2, KBr, toluene (C7H8, a constituent of gasoline).

PROBLEM 11.2 Which would you expect to have the larger (more negative) hydration energy?

(a) Na+ or Cs+ (b) K+ or Ba2+



WORKED EXAMPLE 11.2 Using a Mass Percent Concentration

Strategy

Assume that you have a 5.75 mass % solution of LiCl in water. What mass of solution in grams contains

1.60 g of LiCl?

Describing a concentration as 5.75 mass % means that 100.0 g of solution contains 5.75 g of LiCl (and 94.25 g of

H2O), a relationship that can be used as a conversion factor.

Solution



WORKED EXAMPLE 11.3 Using Density to Convert Mass

Percent Concentration to Molarity

Strategy

The density of a 25.0 mass % solution of sulfuric acid (H2SO4) in water is 1.1783 g/mL at 25.0 C. What is

the molarity of the solution?

Describing a solution as 25.0 mass % sulfuric acid in water means that 100.0 g of solution contains 25.0 g of H2SO4

and 75.0 g of water. Because we want to calculate the concentration in molarity, we first need to find the number of

moles of sulfuric acid dissolved in a specific mass of solution. We next use density as a conversion factor to find the

volume of that solution and then calculate molarity by dividing the number of moles by the volume.

Solution

First, convert the 25.0 g of H2SO4 into moles:

Next, find the volume of 100.0 g of solution, using density as the conversion factor:

Then, calculate the molarity of the solution:

The molarity of the 25.0 mass % sulfuric acid solution is 3.00 M.



WORKED EXAMPLE 11.3 Using Density to Convert Mass

Percent Concentration to MolarityContinued

PROBLEM 11.3 What is the mass percent concentration of a saline solution prepared by dissolving 1.00 mol of

NaCl in 1.00 L of water?

PROBLEM 11.4 The legal limit for human exposure to carbon monoxide in the workplace is 35 ppm. Assuming

that the density of air is 1.3 g/L, how many grams of carbon monoxide are in 1.0 L of air at the maximum

allowable concentration?

PROBLEM 11.5 Assuming that seawater is an aqueous solution of NaCl, what is its molarity? The density of

seawater is 1.025 g/mL at 20 °C, and the NaCl concentration is 3.50 mass %.



WORKED EXAMPLE 11.4 Calculating the Molality of a Solution

Strategy

What is the molality of a solution made by dissolving 1.45 g of table sugar (sucrose, C12H22O11) in 30.0 mL

of water? The molar mass of sucrose is 342.3 g/mol.

Molality is the number of moles of solute per kilogram of solvent. Thus, we need to find how many moles are in

1.45 g of sucrose and how many kilograms are in 30.0 mL of water.

Solution

The number of moles of sucrose is

Since the density of water is 1.00 g/mL, 30.0 mL of water has a mass of 30.0 g, or 0.0300 kg. Thus, the molality of

the solution is



WORKED EXAMPLE 11.5 Using Density to Convert from Molality

to Molarity

Strategy

Ethylene glycol, C2H4(OH)2, is a colorless liquid used as automobile antifreeze. If the density at 20 C of a

4.028 m solution of ethylene glycol in water is 1.0241 g/mL, what is the molarity of the solution? The molar

mass of ethylene glycol is 62.07 g/mol.

A 4.028 m solution of ethylene glycol in water contains 4.028 mol of ethylene glycol per kilogram of water. To find

the solution’s molarity, we need to find the number of moles of solute per volume (liter) of solution. The volume, in

turn, can be found from the mass of the solution by using density as a conversion factor.

Solution

The mass of the solution is the sum of the masses of solute and solvent. Assuming that 1.000 kg of solvent is

used to dissolve 4.028 mol of ethylene glycol, the mass of the ethylene glycol is

Dissolving this 250.0 g of ethylene glycol in 1.000 kg (or 1000 g) of water gives the total mass of the solution:

The volume of the solution is obtained from its mass by using density as a conversion factor:

The molarity of the solution is the number of moles of solute divided by the volume of solution:



WORKED EXAMPLE 11.6 Using Density to Convert Molarity to

Other Measures of ConcentrationA 0.750 M solution of H2SO4 in water has a density of 1.046 g/mL at 20 C. What is the concentration of

this solution in (a) mole fraction, (b) mass percent, and (c) molality? The molar mass of H2SO4 is 98.1

g/mol.

(a) Let’s pick an arbitrary amount of the solution that will make the calculations easy, say 1.00 L. Since the

concentration of the solution is 0.750 mol/L and the density is 1.046 g/mL (or 1.046 kg/L), 1.00 L of the solution

contains 0.750 mol (73.6 g) of H2SO4 and has a mass of 1.046 kg:

Strategy and Solution

Subtracting the mass of H2SO4 from the total mass of the solution gives 0.972 kg, or 54.0 mol, of water in

1.00 L of solution:

Thus, the mole fraction of H2SO4 is



WORKED EXAMPLE 11.6 Using Density to Convert Molarity to

Other Measures of Concentration

(b) The mass percent concentration can be determined from the calculations in part (a):

Thus, the molality of the sulfuric acid solution is 0.772 m.

(c) The molality of the solution can also be determined from the calculations in part (a). Since 0.972 kg of water has

0.750 mol of H2SO4 dissolved in it, 1.00 kg of water would have 0.772 mol of H2SO4 dissolved in it:

Continued

PROBLEM 11.6 What is the molality of a solution prepared by dissolving 0.385 g of cholesterol, C27H46O, in

40.0 g of chloroform, CHCl3? What is the mole fraction of cholesterol in the solution?

PROBLEM 11.7 What mass in grams of a 0.500 m solution of sodium acetate, CH3CO2Na, in water would you

use to obtain 0.150 mol of sodium acetate?

PROBLEM 11.8 The density at 20 °C of a 0.258 m solution of glucose in water is 1.0173 g/mL, and the molar

mass of glucose is 180.2 g. What is the molarity of the solution?

PROBLEM 11.9 The density at 20 °C of a 0.500 M solution of acetic acid in water is 1.0042 g/mL. What is

the molality of the solution? The molar mass of acetic acid, CH3CO2H, is 60.05 g.

PROBLEM 11.10 Assuming that seawater is a 3.50 mass % aqueous solution of NaCl, what is the molality of

seawater?



WORKED EXAMPLE 11.7 Using Henry’s Law to Calculate Gas

SolubilityThe Henry’s-law constant of methyl bromide (CH3Br), a gas used as a soil fumigating agent, is

k = 0.159 mol/(L • atm) at 25 C. What is the solubility in mol/L of methyl bromide in water at 25 C

and a partial pressure of 125 mm Hg?

According to Henry’s law, solubility = k P.

Strategy

The solubility of methyl bromide in water at a partial pressure of 125 mm Hg is 0.0261 M.

Solution

PROBLEM 11.11 The solubility of CO2 in water is 3.2 10–2 M at 25 °C and 1 atm pressure. What is

the Henry’s-law constant for CO2 in mol/(L atm)?

PROBLEM 11.12 The partial pressure of CO2 in air is approximately 4.0 10–4 atm. Use the Henry’s-law

constant you calculated in Problem 11.11 to find the concentration of CO2 in:

(a) A can of soda under a CO2 pressure of 2.5 atm at 25 °C

(b) A can of soda open to the atmosphere at 25 °C



WORKED EXAMPLE 11.8 Calculating the Vapor pressure of a

Solution

What is the vapor pressure in mm Hg of a solution made by dissolving 18.3 g of NaCl in 500.0 g of H2O at

70 C, assuming a van’t Hoff factor of 1.9? The vapor pressure of pure water at 70 C is 233.7 mm Hg.

According to Raoult’s law, the vapor pressure of the solution equals the vapor pressure of pure solvent times the

mole fraction of the solvent in the solution. Thus, we have to find the numbers of moles of solvent and solute and

then calculate the mole fraction of solvent.

Strategy

First, use molar mass to calculate the number of moles of NaCl and H2O.

Solution

Next, calculate the mole fraction of water in the solution. A van’t Hoff factor of 1.9 means that the NaCl

dissociates incompletely and gives only 1.9 particles per formula unit. Thus, the solution contains 1.9 0.313 mol

= 0.59 mol of dissolved particles and the mole fraction of water is

From Raoult’s law, the vapor pressure of the solution is



WORKED EXAMPLE 11.9 Calculating Vapor-pressure Lowering

How many grams of sucrose must be added to 320 g of water to lower the vapor pressure by 1.5 mm Hg at

25 C? The vapor pressure of water at 25 C is 23.8 mm Hg, and the molar mass of sucrose is 342.3 g/mol.

According to Raoult’s law, Psoln = Psolv Xsolv, which can be rearranged to the form Xsolv = Psoln/Psolv. This equation

can then be solved to find the number of moles of sucrose and hence the number of grams.

Strategy

First, calculate the vapor pressure of the solution, Psoln, by subtracting the amount of vapor-pressure lowering from

the vapor pressure of the pure solvent, Psolv:

Solution

Now calculate the mole fraction of water, Xsolv.

This mole fraction of water is the number of moles of water divided by the total number of moles of sucrose

plus water:

Since the number of moles of water is



WORKED EXAMPLE 11.9 Calculating Vapor-pressure Lowering

then the total number of moles of sucrose plus water is

Subtracting the number of moles of water from the total number of moles gives the number of moles of

sucrose needed:

Continued

Converting moles into grams then gives the mass of sucrose needed:

PROBLEM 11.13 What is the vapor pressure in mm Hg of a solution prepared by dissolving 5.00 g of

benzoic acid (C7H6O2) in 100.00 g of ethyl alcohol (C2H6O) at 35 °C? The vapor pressure of pure ethyl

alcohol at 35 °C is 100.5 mm Hg.

PROBLEM 11.14 How many grams of NaBr must be added to 250 g of water to lower the vapor pressure by

1.30 mm Hg at 40 °C assuming complete dissociation? The vapor pressure of water at 40 °C is 55.3 mm Hg.

KEY CONCEPT PROBLEM 11.15 The following diagram shows a close-up view of part of the vapor-

pressure curve for a pure solvent and a solution of a nonvolatile solute. Which curve represents the pure solvent,

and which the solution?



WORKED KEY CONCEPT EXAMPLE 11.10 Interpreting Vapor-

pressure Curves

The following diagram shows a close-up view of part of the vapor-pressure curves for two pure liquids and a

mixture of the two. Which curves represent pure liquids? Which represents the mixture?

The vapor pressure of a mixture of two volatile liquids is always intermediate between the vapor pressures of the

two pure liquids. Thus, the top (red) and bottom (blue) curves represent pure liquids, and the middle curve (green)

represents the mixture.

Strategy and Solution

PROBLEM 11.16

(a) What is the vapor pressure in mm Hg of a solution prepared by dissolving 25.0 g of ethyl alcohol (C2H5OH)

in 100.0 g of water at 25 °C? The vapor pressure of pure water is 23.8 mm Hg, and the vapor pressure of

ethyl alcohol is 61.2 mm Hg at 25 °C.

(b) What is the vapor pressure of the solution if 25.0 g of water is dissolved in 100.0 g of ethyl alcohol at 25 °C?



WORKED KEY CONCEPT EXAMPLE 11.10 Interpreting Vapor-

pressure CurvesContinued

KEY CONCEPT PROBLEM 11.17 The following phase diagram shows part of the vapor-pressure curves for a

pure liquid (green curve) and a solution of the first liquid with a second volatile liquid (red curve).

(a) Is the boiling point of the second liquid higher or lower than that of the first liquid?

(b) Draw on the diagram the approximate position of the vapor-pressure curve for the second liquid.



WORKED EXAMPLE 11.11 Using Boiling-Point Elevation to

Calculate the Molality of a Solution

What is the molality of an aqueous glucose solution if the boiling point of the solution at 1 atm pressure is

101.27 C? The molal boiling-point-elevation constant for water is given in Table 11.4.

Rearrange the equation for molal boiling-point elevation to solve for m:

Strategy

where Kb = 0.51 (C • kg)/mol and Tb = 101.27 C – 100.00 C = 1.27 C.

Solution

The molality of the solution is 2.5 m.




Calculate the Molality of a SolutionContinued

PROBLEM 11.18 What is the normal boiling point in °C of a solution prepared by dissolving 1.50 g of aspirin

(acetylsalicylic acid, C9H8O4) in 75.00 g of chloroform (CHCl3)? The normal boiling point of chloroform is

61.7 °C, and Kb for chloroform is given in Table 11.4.

PROBLEM 11.19 What is the freezing point in °C of a solution prepared by dissolving 7.40 g of MgCl2 in

110 g of water? The value of Kf for water is given in Table 11.4, and the van’t Hoff factor for MgCl2 is i = 2.7.

PROBLEM 11.20 Assuming complete dissociation, what is the molality of an aqueous solution of KBr whose

freezing point is –2.95 °C? The molal freezing-point-depression constant of water is given in Table 11.4.




Calculate the Molality of a SolutionContinued

PROBLEM 11.21 When 9.12 g of HCl was dissolved in 190 g of water, the freezing point of the solution was

–4.65 °C. What is the value of the van’t Hoff factor for HCl?

KEY CONCEPT PROBLEM 11.22 The following phase diagram shows a close-up view of the liquid/vapor

phase transition boundaries for pure chloroform and a solution of a nonvolatile solute in chloroform.

(a) What is the approximate boiling point of pure chloroform?

(b) What is the approximate molal concentration of the nonvolatile solute? See Table 11.4 to find Kb for

chloroform.



WORKED EXAMPLE 11.12 Calculating the Osmotic Pressure of a

SolutionThe total concentration of dissolved particles inside red blood cells is approximately 0.30 M, and the

membrane surrounding the cells is semipermeable. What would the osmotic pressure in atmospheres inside

the cells become if the cells were removed from blood plasma and placed in pure water at 298 K?

If red blood cells were removed from the body and placed in pure water, water would pass through the cell

membrane, causing an increase in pressure inside the cells. The amount of this pressure would be

Strategy

where M = 0.30mol/L, R = 0.082 06 (L • atm)/(K • mol), T = 298 K.

Solution

The buildup of internal pressure would cause the blood cells to burst.



WORKED EXAMPLE 11.13 Using Osmotic Pressure to Calculate

the Molarity of a Solution

A solution of an unknown substance in water at 293 K gives rise to an osmotic pressure of 5.66 atm. What is

the molarity of the solution?

We are given values for and T, and we need to solve for M in the equation = MRT. Rearranging this

equation, we get

Strategy

where = 5.66 atm, R = 0.082 06 (L • atm)/(K • mol), and T = 293 K.

Solution

PROBLEM 11.23 What osmotic pressure in atmospheres would you expect for a solution of 0.125 M CaCl2 that is

separated from pure water by a semipermeable membrane at 310 K? Assume 100% dissociation for CaCl2.

PROBLEM 11.24 A solution of an unknown substance in water at 300 K gives rise to an osmotic pressure of 3.85

atm. What is the molarity of the solution?




the Molecular Mass of a Solute

A solution prepared by dissolving 20.0 mg of insulin in water and diluting to a volume of 5.00 mL gives an

osmotic pressure of 12.5 mm Hg at 300 K. What is the molecular mass of insulin?

To determine molecular mass, we need to know the number of moles of insulin represented by the 20.0 mg

sample. We can do this by first rearranging the equation for osmotic pressure to find the molar concentration of

the insulin solution and then multiplying by the volume of the solution to obtain the number of moles of insulin.

Strategy

Since the volume of the solution is 5.00 mL, the number of moles of insulin is

Solution

Knowing both the mass and the number of moles of insulin, we can calculate the molar mass and hence the

molecular mass:

The molecular mass of insulin is 5990 amu.




the Molecular Mass of a SoluteContinued

PROBLEM 11.25 A solution of 0.250 g of naphthalene (mothballs) in 35.00 g of camphor lowers the freezing

point by 2.10 °C. What is the molar mass of naphthalene? The freezing-point-depression constant for camphor

is 37.7 (°C kg)/mol.

PROBLEM 11.26 What is the molar mass of sucrose (table sugar) if a solution prepared by dissolving

0.822 g of sucrose in 300.0 mL of water has an osmotic pressure of 149 mm Hg at 298 K?

worked example 11.1 correlating solubility and chemical...

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