Solved Examples (Dedicated Problem Set)

How to Use This Problem Set

This is your dedicated workout for the Solutions chapter. The problems below run from warm-up to exam-level, grouped by theme: concentration units, Henry's law, Raoult's law and vapour pressure, the four colligative properties, and the van't Hoff factor.

Work each one with a pen before reading the solution. For every numerical, follow the same discipline you've built up: write the formula first, substitute with units, then compute. Useful constants to keep handy:

• $K_b$ (water) $= 0.52$ K kg mol$^{-1}$; $K_f$ (water) $= 1.86$ K kg mol$^{-1}$.
• $R = 0.0821$ L atm K$^{-1}$ mol$^{-1}$ $= 0.083$ L bar K$^{-1}$ mol$^{-1}$.
• $T(\text{K}) = t(\text{°C}) + 273$.

Let's get to work.

Solved Examples — Concentration

Example 1

Calculate the mole fraction of ethanol (M = 46) in a solution containing 46 g ethanol and 90 g water (M = 18).

Solution: $n_{eth}=46/46=1.0$; $n_{water}=90/18=5.0$. $x_{eth}=\dfrac{1.0}{1.0+5.0}=0.167$. Answer: 0.167.

Example 2

Find the molality of a solution of 9.8 g H$_2$SO$_4$ (M = 98) in 250 g of water.

Solution: $n=9.8/98=0.10$ mol; mass of solvent $=0.250$ kg. $m=0.10/0.250=0.40$ mol kg$^{-1}$. Answer: 0.40 m.

Example 3

A 2.0 M aqueous solution of NaOH (M = 40) has density 1.08 g mL$^{-1}$. Find its molality.

Solution: Take 1 L = 1000 mL → mass $=1000\times1.08=1080$ g. NaOH in 2.0 mol $=80$ g. Water $=1080-80=1000$ g $=1.0$ kg. $m=2.0/1.0=2.0$ m. Answer: 2.0 m.

Example 4

Concentrated HCl is 37% by mass with density 1.19 g mL$^{-1}$ (M = 36.5). Find its molarity.

Solution: In 100 g solution: HCl $=37$ g $=37/36.5=1.014$ mol; volume $=100/1.19=84.0$ mL $=0.0840$ L. $M=1.014/0.0840=12.07$ M. Answer: ≈ 12.1 M.

Example 5

How many grams of glucose (M = 180) are needed to prepare 500 mL of a 0.25 M solution?

Solution: moles $=0.25\times0.500=0.125$ mol; mass $=0.125\times180=22.5$ g. Answer: 22.5 g.

Example 6

Convert a 0.5 molal aqueous glucose solution (M = 180) to mass percentage.

Solution: 0.5 mol glucose $=0.5\times180=90$ g in 1000 g water. Mass % $=\dfrac{90}{90+1000}\times100=8.26\%$. Answer: 8.26%.

Solved Examples — Henry's Law

Example 7

The Henry constant for O$_2$ in water at 298 K is $3.30\times10^7$ torr. If the partial pressure of O$_2$ is 0.21 atm (= 159.6 torr), find the mole fraction of O$_2$ dissolved.

Solution: $x=p/K_H=159.6/(3.30\times10^7)=4.84\times10^{-6}$. Answer: $4.84\times10^{-6}$.

Example 8

Henry's constant for CO$_2$ in water at 298 K is 1.67 × 10$^8$ Pa. Calculate the mole fraction of CO$_2$ in soda water when the partial pressure of CO$_2$ is 2.5 atm. (1 atm = 1.013 × 10$^5$ Pa)

Solution: $p=2.5\times1.013\times10^5=2.533\times10^5$ Pa. $x=p/K_H=2.533\times10^5/1.67\times10^8=1.517\times10^{-3}$. Answer: ≈ $1.52\times10^{-3}$.

Example 9

Two gases X ($K_H=8$ kbar) and Y ($K_H=35$ kbar) are in water. At the same partial pressure, find the ratio of their dissolved mole fractions $x_X:x_Y$.

Solution: $x=p/K_H$, so $x_X:x_Y=\dfrac{1}{8}:\dfrac{1}{35}=35:8$. Answer: $35:8$ (X is more soluble).

Solved Examples — Raoult's Law & Vapour Pressure

Example 10

Liquids A ($p^0_A=120$ mmHg) and B ($p^0_B=180$ mmHg) form an ideal solution with $x_A=0.3$. Find $p_{total}$.

Solution: $p=120(0.3)+180(0.7)=36+126=162$ mmHg. Answer: 162 mmHg.

Example 11

For the solution in Example 10, find the mole fraction of A in the vapour.

Solution: $y_A=p_A/p_{total}=36/162=0.222$. Answer: 0.222 (vapour is richer in the more volatile B, as expected).

Example 12

The vapour pressure of pure water at 298 K is 23.8 mmHg. Calculate the vapour pressure of a solution containing 5.85 g of a non-volatile non-electrolyte (M = 58.5) in 180 g water (M = 18).

Solution: $n_{solute}=5.85/58.5=0.10$; $n_{water}=180/18=10.0$; $x_{water}=10.0/10.1=0.990$. $p=23.8\times0.990=23.56$ mmHg. Answer: 23.56 mmHg.

Example 13

A solution of two volatile liquids has $p^0_A=100$, $p^0_B=300$ mmHg. If the vapour contains equal moles of A and B ($y_A=y_B=0.5$), find $x_A$ in the liquid.

Solution: $y_A=\dfrac{p^0_A x_A}{p^0_A x_A+p^0_B(1-x_A)}$. Set $=0.5$: $100x_A=300(1-x_A)\Rightarrow100x_A=300-300x_A\Rightarrow400x_A=300\Rightarrow x_A=0.75$. Answer: $x_A=0.75$.

Solved Examples — Relative Lowering & Molar Mass

Example 14

The vapour pressure of pure benzene is 0.850 bar; adding 5 g of a non-volatile solute to 100 g benzene (M = 78) gives 0.83 bar. Find the molar mass of the solute.

Solution: Relative lowering $=\dfrac{0.850-0.83}{0.850}=0.02353$. $n_{benzene}=100/78=1.282$. $n_{solute}\approx0.02353\times1.282=0.03017$ mol. $M=5/0.03017=165.7$ g mol$^{-1}$. Answer: ≈ 166 g mol$^{-1}$.

Example 15

18 g of glucose (M = 180) is dissolved in 1 kg of water. What is the vapour pressure lowering relative to pure water at 100 °C (where $p^0=760$ mmHg)?

Solution: $n_{glu}=0.10$; $n_{water}=1000/18=55.55$. $x_{solute}=\dfrac{0.10}{55.65}=1.797\times10^{-3}$. Lowering $=p^0 x_{solute}=760\times1.797\times10^{-3}=1.366$ mmHg. Answer: ≈ 1.37 mmHg.

Example 16

The relative lowering of vapour pressure of a dilute solution is 0.008. If the solvent is water (M = 18) and 200 g of it is used, how many moles of solute are present?

Solution: $n_A=200/18=11.11$ mol. $n_B\approx0.008\times11.11=0.0889$ mol. Answer: ≈ 0.089 mol.

Solved Examples — Boiling & Freezing Point

Example 17

A solution of 3.0 g of a non-electrolyte in 100 g of water boils at 100.13 °C. Find the molar mass. ($K_b=0.52$)

Solution: $\Delta T_b=0.13$ K. $M=\dfrac{K_b w_B 1000}{\Delta T_b w_A}=\dfrac{0.52\times3.0\times1000}{0.13\times100}=\dfrac{1560}{13}=120$ g mol$^{-1}$. Answer: 120 g mol$^{-1}$.

Example 18

What mass of ethylene glycol (M = 62) must be added to 5.0 kg of water to lower its freezing point to −4.0 °C? ($K_f=1.86$)

Solution: $\Delta T_f=4.0$ K. $m=\Delta T_f/K_f=4.0/1.86=2.151$ mol kg$^{-1}$. moles $=2.151\times5.0=10.75$ mol. mass $=10.75\times62=666.7$ g. Answer: ≈ 667 g.

Example 19

A 5% (by mass) aqueous solution of a non-electrolyte (M = 90) — find its freezing point. ($K_f=1.86$; assume mass of solution ≈ mass of water for dilute)

Solution: In 100 g solution: 5 g solute, 95 g water. $m=\dfrac{5/90}{0.095}=\dfrac{0.0556}{0.095}=0.585$ m. $\Delta T_f=1.86\times0.585=1.088$ K. f.p. $=-1.09$ °C. Answer: ≈ −1.09 °C.

Solved Examples — Additional Colligative Property Practice

Example 20

The boiling point of a solution made by dissolving 12.0 g of a solute in 200 g water is 100.26 °C. Find the solute's molar mass. ($K_b=0.52$)

Solution: $\Delta T_b=0.26$ K. $M=\dfrac{0.52\times12.0\times1000}{0.26\times200}=\dfrac{6240}{52}=120$ g mol$^{-1}$. Answer: 120 g mol$^{-1}$.

Example 21

Camphor ($K_f=40$ K kg mol$^{-1}$) is an excellent cryoscopic solvent. 0.5 g of a solute in 25 g of camphor lowers the freezing point by 8.0 K. Find the solute's molar mass.

Solution: $M=\dfrac{K_f w_B 1000}{\Delta T_f w_A}=\dfrac{40\times0.5\times1000}{8.0\times25}=\dfrac{20000}{200}=100$ g mol$^{-1}$. Answer: 100 g mol$^{-1}$.

Example 22

A solution containing 8 g of a substance in 100 g of diethyl ether ($K_b=2.02$) boils 0.50 K higher than pure ether. Find the molar mass.

Solution: $M=\dfrac{2.02\times8\times1000}{0.50\times100}=\dfrac{16160}{50}=323.2$ g mol$^{-1}$. Answer: ≈ 323 g mol$^{-1}$.

Solved Examples — Osmotic Pressure

Example 23

Calculate the osmotic pressure of a solution containing 18 g glucose (M = 180) per litre at 300 K. ($R=0.0821$)

Solution: $C=\dfrac{18/180}{1}=0.10$ M. $\Pi=CRT=0.10\times0.0821\times300=2.46$ atm. Answer: 2.46 atm.

Example 24

A 5% (w/v) solution of cane sugar (M = 342) — find its osmotic pressure at 300 K. ($R=0.0821$)

Solution: 5% w/v = 5 g per 100 mL = 50 g L$^{-1}$. $C=50/342=0.1462$ M. $\Pi=0.1462\times0.0821\times300=3.60$ atm. Answer: ≈ 3.60 atm.

Example 25

The osmotic pressure of a 0.0019 M sucrose solution at 300 K matches that of a protein solution containing 1.0 g protein in 200 mL. Find the protein's molar mass. ($R=0.0821$; use $\Pi$ of sucrose)

Solution: $\Pi=0.0019\times0.0821\times300=0.0468$ atm. $M=\dfrac{w R T}{\Pi V}=\dfrac{1.0\times0.0821\times300}{0.0468\times0.200}=\dfrac{24.63}{0.00936}=2631$… Let's compute: $=2.63\times10^3$ g mol$^{-1}$. Answer: ≈ 2.6 × 10$^3$ g mol$^{-1}$.

Solved Examples — Osmotic Pressure Practice

Example 26

What concentration of glucose solution would be isotonic with a 0.05 M urea solution at the same temperature?

Solution: Isotonic → equal molarity (both $i=1$): $C_{glucose}=0.05$ M. Answer: 0.05 M.

Example 27

An aqueous solution of a non-electrolyte has osmotic pressure 4.92 atm at 300 K. Find its molarity. ($R=0.0821$)

Solution: $C=\Pi/RT=4.92/(0.0821\times300)=4.92/24.63=0.20$ M. Answer: 0.20 M.

Solved Examples — van't Hoff Factor

Example 28

The freezing point depression of a 0.2 m aqueous KCl solution is 0.68 K. Find the van't Hoff factor. ($K_f=1.86$)

Solution: Normal $\Delta T_f=K_f m=1.86\times0.2=0.372$ K. $i=\dfrac{0.68}{0.372}=1.83$. Answer: $i=1.83$.

Example 29

For Example 28, find the degree of dissociation of KCl. (KCl → 2 ions)

Solution: $i=1+(n-1)\alpha=1+\alpha$ (since $n=2$). $1.83=1+\alpha\Rightarrow\alpha=0.83$. Answer: $\alpha=0.83$ (83%).

Example 30

Calculate the osmotic pressure of a 0.05 M K$_2$SO$_4$ solution at 300 K, assuming complete dissociation. ($R=0.0821$)

Solution: K$_2$SO$_4$ → 3 ions, $i=3$. $\Pi=iCRT=3\times0.05\times0.0821\times300=3.69$ atm. Answer: ≈ 3.69 atm.

Solved Examples — van't Hoff Factor Practice

Example 31

Acetic acid (M = 60) dimerises in benzene. A 1.0 m solution gives a freezing-point depression corresponding to an apparent molality of 0.6 m. Find the van't Hoff factor and degree of association.

Solution: $i=\dfrac{\text{apparent}}{\text{actual}}=\dfrac{0.6}{1.0}=0.6$. For dimerisation $i=1-\dfrac{\alpha}{2}$: $0.6=1-\dfrac{\alpha}{2}\Rightarrow\alpha=0.8$. Answer: $i=0.6$, $\alpha=0.8$ (80% associated).

Example 32

A 0.1 M solution of Al$_2$(SO$_4$)$_3$ is assumed completely dissociated. Find its van't Hoff factor and the effective particle concentration.

Solution: Al$_2$(SO$_4$)$_3$ → 2Al$^{3+}$ + 3SO$_4^{2-}$ = 5 ions, so $i=5$. Effective concentration $=i\times C=5\times0.1=0.5$ M of particles. Answer: $i=5$; effective particle concentration $=0.5$ M.

Example 33

Which has the higher boiling point: 0.1 m glucose or 0.1 m NaCl (complete dissociation)? Justify with numbers. ($K_b=0.52$)

Solution: glucose ($i=1$): $\Delta T_b=0.52\times0.1=0.052$ K. NaCl ($i=2$): $\Delta T_b=2\times0.52\times0.1=0.104$ K. NaCl gives the larger elevation, hence the higher boiling point. Answer: 0.1 m NaCl (twice the elevation of glucose).