Chapter 1: Solutions

Consider a 100 g sample of the solution. Since it is 20% $$\mathrm{C_2H_6O_2}$$ by mass:

Mass of ethylene glycol $$= 20 \, \mathrm{g}$$
Mass of water $$= 100 - 20 = 80 \, \mathrm{g}$$

Molar mass of $$\mathrm{C_2H_6O_2} = 2(12) + 6(1) + 2(16) = 62 \, \mathrm{g \, mol^{-1}}$$

Number of moles of ethylene glycol:

$$n_{\text{glycol}} = \dfrac{20}{62} = 0.322 \, \mathrm{mol}$$

Number of moles of water:

$$n_{\text{water}} = \dfrac{80}{18} = 4.444 \, \mathrm{mol}$$

The mole fraction of ethylene glycol equals its moles divided by the total moles of all components:

$$x_{\text{glycol}} = \dfrac{n_{\text{glycol}}}{n_{\text{glycol}} + n_{\text{water}}} = \dfrac{0.322}{0.322 + 4.444} = \dfrac{0.322}{4.766}$$

$$x_{\text{glycol}} = 0.068$$

Molarity is defined as the number of moles of solute present in one litre of solution.

Molar mass of NaOH $$= 23 + 16 + 1 = 40 \, \mathrm{g \, mol^{-1}}$$

Number of moles of NaOH:

$$n_{\text{NaOH}} = \dfrac{5 \, \mathrm{g}}{40 \, \mathrm{g \, mol^{-1}}} = 0.125 \, \mathrm{mol}$$

Volume of the solution $$= 450 \, \mathrm{mL} = 0.450 \, \mathrm{L}$$

Therefore,

$$\text{Molarity} = \dfrac{n_{\text{NaOH}}}{V_{\text{solution}} \,(\mathrm{L})} = \dfrac{0.125 \, \mathrm{mol}}{0.450 \, \mathrm{L}}$$

$$\text{Molarity} = 0.278 \, \mathrm{M}$$

Molality is defined as the number of moles of solute present in one kilogram of solvent.

Molar mass of $$\mathrm{CH_3COOH} = 2(12) + 4(1) + 2(16) = 60 \, \mathrm{g \, mol^{-1}}$$

Number of moles of ethanoic acid:

$$n = \dfrac{2.5 \, \mathrm{g}}{60 \, \mathrm{g \, mol^{-1}}} = 0.0417 \, \mathrm{mol}$$

Mass of benzene (solvent) $$= 75 \, \mathrm{g} = 0.075 \, \mathrm{kg}$$

Therefore,

$$\text{Molality} = \dfrac{0.0417 \, \mathrm{mol}}{0.075 \, \mathrm{kg}}$$

$$\text{Molality} = 0.556 \, \mathrm{mol \, kg^{-1}}$$

The mass percentage of a component is its mass divided by the total mass of the solution, multiplied by 100.

Total mass of the solution:

$$m_{\text{total}} = 22 + 122 = 144 \, \mathrm{g}$$

Mass percentage of benzene:

$$\%\,\mathrm{C_6H_6} = \dfrac{22}{144} \times 100 = 15.28\,\%$$

Mass percentage of carbon tetrachloride:

$$\%\,\mathrm{CCl_4} = \dfrac{122}{144} \times 100 = 84.72\,\%$$

Check: $$15.28\,\% + 84.72\,\% = 100\,\%$$, as expected.

Consider 100 g of the solution. It contains 30 g of benzene and $$100 - 30 = 70 \, \mathrm{g}$$ of carbon tetrachloride.

Molar masses:

$$\mathrm{C_6H_6}: \; 6(12) + 6(1) = 78 \, \mathrm{g \, mol^{-1}}$$
$$\mathrm{CCl_4}: \; 12 + 4(35.5) = 154 \, \mathrm{g \, mol^{-1}}$$

Number of moles of benzene:

$$n_{\mathrm{C_6H_6}} = \dfrac{30}{78} = 0.3846 \, \mathrm{mol}$$

Number of moles of carbon tetrachloride:

$$n_{\mathrm{CCl_4}} = \dfrac{70}{154} = 0.4545 \, \mathrm{mol}$$

Mole fraction of benzene:

$$x_{\mathrm{C_6H_6}} = \dfrac{n_{\mathrm{C_6H_6}}}{n_{\mathrm{C_6H_6}} + n_{\mathrm{CCl_4}}} = \dfrac{0.3846}{0.3846 + 0.4545} = \dfrac{0.3846}{0.8391}$$

$$x_{\mathrm{C_6H_6}} = 0.458$$

(a) 30 g of $$\mathrm{Co(NO_3)_2 \cdot 6H_2O}$$ in 4.3 L of solution

Molar mass of $$\mathrm{Co(NO_3)_2 \cdot 6H_2O}$$:

$$= 59 + 2(14 + 3 \times 16) + 6(2 \times 1 + 16)$$
$$= 59 + 2(62) + 6(18) = 59 + 124 + 108 = 291 \, \mathrm{g \, mol^{-1}}$$

Number of moles of the solute:

$$n = \dfrac{30}{291} = 0.103 \, \mathrm{mol}$$

Molarity:

$$M = \dfrac{0.103 \, \mathrm{mol}}{4.3 \, \mathrm{L}} = 0.024 \, \mathrm{M}$$

(b) 30 mL of 0.5 M $$\mathrm{H_2SO_4}$$ diluted to 500 mL

On dilution the number of moles of solute does not change, so the dilution relation applies:

$$M_1 V_1 = M_2 V_2$$

$$M_2 = \dfrac{M_1 V_1}{V_2} = \dfrac{0.5 \, \mathrm{M} \times 30 \, \mathrm{mL}}{500 \, \mathrm{mL}} = \dfrac{15}{500}$$

$$M_2 = 0.03 \, \mathrm{M}$$

A 0.25 molal solution contains 0.25 mol of urea per 1 kg (1000 g) of water.

Molar mass of urea $$\mathrm{(NH_2CONH_2)} = 2(14) + 4(1) + 12 + 16 = 60 \, \mathrm{g \, mol^{-1}}$$

Let the mass of urea required be $$w$$ grams. The total mass of the solution is $$2.5 \, \mathrm{kg} = 2500 \, \mathrm{g}$$, so the mass of water is $$(2500 - w)$$ grams.

Number of moles of urea $$= \dfrac{w}{60}$$.

Applying the definition of molality:

$$\text{molality} = \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}$$

$$0.25 = \dfrac{w/60}{(2500 - w)/1000}$$

$$0.25 = \dfrac{1000\,w}{60\,(2500 - w)}$$

$$0.25 \times 60 \,(2500 - w) = 1000\,w$$

$$15\,(2500 - w) = 1000\,w$$

$$37500 - 15w = 1000w$$

$$37500 = 1015\,w$$

$$w = \dfrac{37500}{1015} = 36.95 \, \mathrm{g}$$

$$w \approx 37 \, \mathrm{g}$$

Consider 100 g of the solution. Being 20% (m/m) KI, it contains 20 g of KI and $$100 - 20 = 80 \, \mathrm{g}$$ of water.

Molar mass of KI $$= 39 + 127 = 166 \, \mathrm{g \, mol^{-1}}$$

Number of moles of KI:

$$n_{\mathrm{KI}} = \dfrac{20}{166} = 0.1205 \, \mathrm{mol}$$

Number of moles of water:

$$n_{\text{water}} = \dfrac{80}{18} = 4.444 \, \mathrm{mol}$$

(a) Molality

$$m = \dfrac{n_{\mathrm{KI}}}{\text{mass of water in kg}} = \dfrac{0.1205}{0.080} = 1.51 \, \mathrm{m}$$

(b) Molarity

Volume of the solution from its density:

$$V = \dfrac{\text{mass}}{\text{density}} = \dfrac{100 \, \mathrm{g}}{1.202 \, \mathrm{g \, mL^{-1}}} = 83.19 \, \mathrm{mL} = 0.08319 \, \mathrm{L}$$

$$M = \dfrac{n_{\mathrm{KI}}}{V} = \dfrac{0.1205}{0.08319} = 1.45 \, \mathrm{M}$$

(c) Mole fraction of KI

$$x_{\mathrm{KI}} = \dfrac{n_{\mathrm{KI}}}{n_{\mathrm{KI}} + n_{\text{water}}} = \dfrac{0.1205}{0.1205 + 4.444} = \dfrac{0.1205}{4.565}$$

$$x_{\mathrm{KI}} = 0.0264$$

By Henry's law, the mole fraction of a dissolved gas is proportional to its partial pressure: $$p = K_H \, x$$.

Given $$p = 0.987 \, \mathrm{bar}$$ and $$K_H = 76.48 \, \mathrm{kbar} = 76.48 \times 10^3 \, \mathrm{bar}$$.

Mole fraction of $$\mathrm{N_2}$$ dissolved in water:

$$x_{\mathrm{N_2}} = \dfrac{p}{K_H} = \dfrac{0.987}{76.48 \times 10^3} = 1.29 \times 10^{-5}$$

1 litre of water has a mass of about 1000 g, so the moles of water are:

$$n_{\text{water}} = \dfrac{1000}{18} = 55.56 \, \mathrm{mol}$$

Since only a tiny amount of gas dissolves, $$x_{\mathrm{N_2}} = \dfrac{n_{\mathrm{N_2}}}{n_{\mathrm{N_2}} + n_{\text{water}}} \approx \dfrac{n_{\mathrm{N_2}}}{n_{\text{water}}}$$.

Therefore the moles of $$\mathrm{N_2}$$ dissolved are:

$$n_{\mathrm{N_2}} = x_{\mathrm{N_2}} \times n_{\text{water}} = 1.29 \times 10^{-5} \times 55.56$$

$$n_{\mathrm{N_2}} = 7.16 \times 10^{-4} \, \mathrm{mol} = 0.716 \, \mathrm{mmol}$$

A solubility of 0.195 m means that 0.195 mol of $$\mathrm{H_2S}$$ is dissolved in 1 kg (1000 g) of water.

Number of moles of water:

$$n_{\text{water}} = \dfrac{1000}{18} = 55.56 \, \mathrm{mol}$$

Mole fraction of $$\mathrm{H_2S}$$ in the solution:

$$x_{\mathrm{H_2S}} = \dfrac{0.195}{0.195 + 55.56} = \dfrac{0.195}{55.75} = 3.50 \times 10^{-3}$$

At STP the partial pressure of the gas is $$p = 0.987 \, \mathrm{bar}$$.

By Henry's law, $$p = K_H \, x$$, so:

$$K_H = \dfrac{p}{x_{\mathrm{H_2S}}} = \dfrac{0.987}{3.50 \times 10^{-3}}$$

$$K_H = 282 \, \mathrm{bar}$$

By Henry's law the mole fraction of dissolved gas is $$x = \dfrac{p}{K_H}$$.

First convert the pressure to pascals:

$$p = 2.5 \, \mathrm{atm} = 2.5 \times 1.013 \times 10^5 \, \mathrm{Pa} = 2.533 \times 10^5 \, \mathrm{Pa}$$

Mole fraction of $$\mathrm{CO_2}$$ in the soda water:

$$x_{\mathrm{CO_2}} = \dfrac{p}{K_H} = \dfrac{2.533 \times 10^5}{1.67 \times 10^8} = 1.517 \times 10^{-3}$$

500 mL of water has a mass of about 500 g, so the moles of water are:

$$n_{\text{water}} = \dfrac{500}{18} = 27.78 \, \mathrm{mol}$$

Since only a small amount of gas dissolves, $$x_{\mathrm{CO_2}} \approx \dfrac{n_{\mathrm{CO_2}}}{n_{\text{water}}}$$. Hence:

$$n_{\mathrm{CO_2}} = x_{\mathrm{CO_2}} \times n_{\text{water}} = 1.517 \times 10^{-3} \times 27.78 = 0.0421 \, \mathrm{mol}$$

Mass of $$\mathrm{CO_2}$$ (molar mass $$44 \, \mathrm{g \, mol^{-1}}$$):

$$w_{\mathrm{CO_2}} = 0.0421 \times 44 = 1.85 \, \mathrm{g}$$

(i) Vapour pressure of the solution

Molar masses:

$$\mathrm{CHCl_3}: 12 + 1 + 3(35.5) = 119.5 \, \mathrm{g \, mol^{-1}}$$
$$\mathrm{CH_2Cl_2}: 12 + 2(1) + 2(35.5) = 85 \, \mathrm{g \, mol^{-1}}$$

Number of moles of each component:

$$n_{\mathrm{CHCl_3}} = \dfrac{25.5}{119.5} = 0.213 \, \mathrm{mol}$$
$$n_{\mathrm{CH_2Cl_2}} = \dfrac{40}{85} = 0.470 \, \mathrm{mol}$$

Mole fractions in the liquid:

$$x_{\mathrm{CHCl_3}} = \dfrac{0.213}{0.213 + 0.470} = \dfrac{0.213}{0.683} = 0.312$$

$$x_{\mathrm{CH_2Cl_2}} = 1 - 0.312 = 0.688$$

By Raoult's law, the total vapour pressure is the sum of the partial pressures:

$$p_{\text{total}} = p^{\circ}_{\mathrm{CHCl_3}} \, x_{\mathrm{CHCl_3}} + p^{\circ}_{\mathrm{CH_2Cl_2}} \, x_{\mathrm{CH_2Cl_2}}$$

$$p_{\text{total}} = (200)(0.312) + (415)(0.688) = 62.4 + 285.5$$

$$p_{\text{total}} = 347.9 \, \mathrm{mm \, Hg}$$

(ii) Mole fractions in the vapour phase

The mole fraction of a component in the vapour equals its partial pressure divided by the total pressure:

$$y_{\mathrm{CHCl_3}} = \dfrac{p_{\mathrm{CHCl_3}}}{p_{\text{total}}} = \dfrac{62.4}{347.9} = 0.18$$

$$y_{\mathrm{CH_2Cl_2}} = \dfrac{p_{\mathrm{CH_2Cl_2}}}{p_{\text{total}}} = \dfrac{285.5}{347.9} = 0.82$$

Let the mole fraction of A in the liquid mixture be $$x_A$$; then the mole fraction of B is $$x_B = 1 - x_A$$.

By Raoult's law, the total vapour pressure is:

$$p_{\text{total}} = p^{\circ}_A \, x_A + p^{\circ}_B \,(1 - x_A)$$

Substituting the given values:

$$600 = 450\,x_A + 700\,(1 - x_A)$$

$$600 = 450\,x_A + 700 - 700\,x_A$$

$$600 - 700 = -250\,x_A$$

$$x_A = \dfrac{100}{250} = 0.40, \qquad x_B = 1 - 0.40 = 0.60$$

Partial pressures of the two components:

$$p_A = p^{\circ}_A \, x_A = 450 \times 0.40 = 180 \, \mathrm{mm \, Hg}$$

$$p_B = p^{\circ}_B \, x_B = 700 \times 0.60 = 420 \, \mathrm{mm \, Hg}$$

Composition of the vapour phase (mole fraction = partial pressure / total pressure):

$$y_A = \dfrac{p_A}{p_{\text{total}}} = \dfrac{180}{600} = 0.30$$

$$y_B = \dfrac{p_B}{p_{\text{total}}} = \dfrac{420}{600} = 0.70$$

For a dilute solution of a non-volatile solute, Raoult's law gives the relative lowering of vapour pressure as the mole fraction of the solute:

$$\dfrac{p^{\circ} - p}{p^{\circ}} = \dfrac{n_2}{n_1 + n_2} \approx \dfrac{n_2}{n_1}$$

where subscript 1 denotes the solvent (benzene) and 2 the solute. Writing the moles in terms of masses and molar masses:

$$\dfrac{p^{\circ} - p}{p^{\circ}} = \dfrac{w_2 / M_2}{w_1 / M_1} = \dfrac{w_2 \, M_1}{M_2 \, w_1}$$

Rearranging for the molar mass of the solute:

$$M_2 = \dfrac{w_2 \, M_1 \, p^{\circ}}{(p^{\circ} - p)\, w_1}$$

Here $$w_2 = 0.5 \, \mathrm{g}$$, $$M_1 = 78 \, \mathrm{g \, mol^{-1}}$$, $$w_1 = 39.0 \, \mathrm{g}$$, $$p^{\circ} = 0.850 \, \mathrm{bar}$$ and $$p^{\circ} - p = 0.850 - 0.845 = 0.005 \, \mathrm{bar}$$.

$$M_2 = \dfrac{0.5 \times 78 \times 0.850}{0.005 \times 39.0} = \dfrac{33.15}{0.195}$$

$$M_2 = 170 \, \mathrm{g \, mol^{-1}}$$

Molar mass of glucose $$\mathrm{C_6H_{12}O_6} = 6(12) + 12(1) + 6(16) = 180 \, \mathrm{g \, mol^{-1}}$$

Number of moles of glucose:

$$n = \dfrac{18}{180} = 0.1 \, \mathrm{mol}$$

Since the glucose is dissolved in 1 kg of water, the molality of the solution is:

$$m = \dfrac{0.1 \, \mathrm{mol}}{1 \, \mathrm{kg}} = 0.1 \, \mathrm{mol \, kg^{-1}}$$

Elevation of boiling point:

$$\Delta T_b = K_b \, m = 0.52 \times 0.1 = 0.052 \, \mathrm{K}$$

Water boils at 373.15 K (100 °C) at 1.013 bar. The solution therefore boils at:

$$T_b = 373.15 + 0.052 = 373.202 \, \mathrm{K}$$

Elevation of the boiling point:

$$\Delta T_b = 354.11 - 353.23 = 0.88 \, \mathrm{K}$$

For a dilute solution $$\Delta T_b = K_b \, m$$, and the molality is

$$m = \dfrac{w_2 \times 1000}{M_2 \, w_1}$$

with $$w_2$$ and $$w_1$$ in grams. Combining the two relations:

$$\Delta T_b = \dfrac{K_b \, w_2 \times 1000}{M_2 \, w_1}$$

Solving for the molar mass of the solute:

$$M_2 = \dfrac{K_b \, w_2 \times 1000}{\Delta T_b \, w_1}$$

Substituting $$K_b = 2.53 \, \mathrm{K \, kg \, mol^{-1}}$$, $$w_2 = 1.80 \, \mathrm{g}$$, $$w_1 = 90 \, \mathrm{g}$$ and $$\Delta T_b = 0.88 \, \mathrm{K}$$:

$$M_2 = \dfrac{2.53 \times 1.80 \times 1000}{0.88 \times 90} = \dfrac{4554}{79.2}$$

$$M_2 \approx 58 \, \mathrm{g \, mol^{-1}}$$

Molar mass of ethylene glycol $$\mathrm{C_2H_6O_2} = 2(12) + 6(1) + 2(16) = 62 \, \mathrm{g \, mol^{-1}}$$

Number of moles of ethylene glycol:

$$n = \dfrac{45}{62} = 0.726 \, \mathrm{mol}$$

The mass of water is $$600 \, \mathrm{g} = 0.600 \, \mathrm{kg}$$, so the molality is:

$$m = \dfrac{0.726 \, \mathrm{mol}}{0.600 \, \mathrm{kg}} = 1.21 \, \mathrm{mol \, kg^{-1}}$$

(a) Freezing point depression

Taking $$K_f = 1.86 \, \mathrm{K \, kg \, mol^{-1}}$$ for water:

$$\Delta T_f = K_f \, m = 1.86 \times 1.21 = 2.25 \, \mathrm{K}$$

(b) Freezing point of the solution

The freezing point of pure water is 273.15 K, so the solution freezes at:

$$T_f = 273.15 - 2.25 = 270.90 \, \mathrm{K}$$

For a dilute solution, the depression of freezing point is $$\Delta T_f = K_f \, m$$, where the molality is

$$m = \dfrac{w_2 \times 1000}{M_2 \, w_1}$$

with masses in grams. Combining the two relations:

$$\Delta T_f = \dfrac{K_f \, w_2 \times 1000}{M_2 \, w_1}$$

Solving for the molar mass of the solute:

$$M_2 = \dfrac{K_f \, w_2 \times 1000}{\Delta T_f \, w_1}$$

Substituting $$K_f = 5.12 \, \mathrm{K \, kg \, mol^{-1}}$$, $$w_2 = 1.00 \, \mathrm{g}$$, $$w_1 = 50 \, \mathrm{g}$$ and $$\Delta T_f = 0.40 \, \mathrm{K}$$:

$$M_2 = \dfrac{5.12 \times 1.00 \times 1000}{0.40 \times 50} = \dfrac{5120}{20}$$

$$M_2 = 256 \, \mathrm{g \, mol^{-1}}$$

The osmotic pressure is related to the molar concentration by $$\pi = C R T$$, where $$C = \dfrac{n}{V} = \dfrac{w}{M \, V}$$.

Therefore:

$$\pi = \dfrac{w \, R T}{M \, V} \quad\Rightarrow\quad M = \dfrac{w \, R T}{\pi \, V}$$

Data: $$w = 1.26 \, \mathrm{g}$$, $$R = 0.083 \, \mathrm{L \, bar \, K^{-1} \, mol^{-1}}$$, $$T = 300 \, \mathrm{K}$$, $$\pi = 2.57 \times 10^{-3} \, \mathrm{bar}$$ and $$V = 200 \, \mathrm{cm^3} = 0.200 \, \mathrm{L}$$.

$$M = \dfrac{1.26 \times 0.083 \times 300}{(2.57 \times 10^{-3})(0.200)}$$

$$M = \dfrac{31.374}{5.14 \times 10^{-4}}$$

$$M \approx 6.1 \times 10^{4} \, \mathrm{g \, mol^{-1}}$$

Molar mass of urea $$\mathrm{(NH_2CONH_2)} = 60 \, \mathrm{g \, mol^{-1}}$$ and molar mass of water $$= 18 \, \mathrm{g \, mol^{-1}}$$.

Number of moles:

$$n_{\text{urea}} = \dfrac{50}{60} = 0.833 \, \mathrm{mol}$$
$$n_{\text{water}} = \dfrac{850}{18} = 47.22 \, \mathrm{mol}$$

Mole fraction of urea (the solute):

$$x_{\text{urea}} = \dfrac{0.833}{0.833 + 47.22} = \dfrac{0.833}{48.06} = 0.0173$$

By Raoult's law, the relative lowering of vapour pressure equals the mole fraction of the solute:

$$\dfrac{p^{\circ} - p}{p^{\circ}} = x_{\text{urea}} = 0.0173$$

So the relative lowering of vapour pressure is $$0.0173$$.

The actual lowering of vapour pressure is:

$$p^{\circ} - p = 0.0173 \times 23.8 = 0.412 \, \mathrm{mm \, Hg}$$

Hence the vapour pressure of water above the solution is:

$$p = 23.8 - 0.412 = 23.4 \, \mathrm{mm \, Hg}$$

The boiling point must be raised from 99.63 °C to 100 °C, so the required elevation is:

$$\Delta T_b = 100 - 99.63 = 0.37\ ^{\circ}\mathrm{C} = 0.37 \, \mathrm{K}$$

For water $$K_b = 0.52 \, \mathrm{K \, kg \, mol^{-1}}$$. From $$\Delta T_b = K_b \, m$$, the molality required is:

$$m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.37}{0.52} = 0.7115 \, \mathrm{mol \, kg^{-1}}$$

The mass of water is $$500 \, \mathrm{g} = 0.500 \, \mathrm{kg}$$, so the moles of sucrose required are:

$$n = m \times \text{mass of water (kg)} = 0.7115 \times 0.500 = 0.3558 \, \mathrm{mol}$$

Molar mass of sucrose $$\mathrm{C_{12}H_{22}O_{11}} = 12(12) + 22(1) + 11(16) = 342 \, \mathrm{g \, mol^{-1}}$$

Mass of sucrose required:

$$w = n \times M = 0.3558 \times 342 = 121.7 \, \mathrm{g}$$

The melting (freezing) point is to be lowered by $$\Delta T_f = 1.5\ ^{\circ}\mathrm{C} = 1.5 \, \mathrm{K}$$.

From $$\Delta T_f = K_f \, m$$, the required molality is:

$$m = \dfrac{\Delta T_f}{K_f} = \dfrac{1.5}{3.9} = 0.3846 \, \mathrm{mol \, kg^{-1}}$$

The mass of acetic acid (the solvent) is $$75 \, \mathrm{g} = 0.075 \, \mathrm{kg}$$, so the moles of ascorbic acid needed are:

$$n = m \times 0.075 = 0.3846 \times 0.075 = 0.02885 \, \mathrm{mol}$$

Molar mass of ascorbic acid $$\mathrm{C_6H_8O_6} = 6(12) + 8(1) + 6(16) = 176 \, \mathrm{g \, mol^{-1}}$$

Mass of ascorbic acid required:

$$w = n \times M = 0.02885 \times 176 = 5.08 \, \mathrm{g}$$

The osmotic pressure is given by $$\pi = \dfrac{n}{V} R T = \dfrac{w}{M \, V} R T$$.

Since the answer is wanted in pascals, work in SI units: $$R = 8.314 \, \mathrm{J \, K^{-1} \, mol^{-1}}$$ and volume in $$\mathrm{m^3}$$.

Number of moles of the polymer:

$$n = \dfrac{w}{M} = \dfrac{1.0}{185000} = 5.41 \times 10^{-6} \, \mathrm{mol}$$

Temperature: $$T = 37\ ^{\circ}\mathrm{C} = 37 + 273 = 310 \, \mathrm{K}$$.

Volume: $$V = 450 \, \mathrm{mL} = 450 \times 10^{-6} \, \mathrm{m^3} = 4.5 \times 10^{-4} \, \mathrm{m^3}$$.

$$\pi = \dfrac{n R T}{V} = \dfrac{(5.41 \times 10^{-6})(8.314)(310)}{4.5 \times 10^{-4}}$$

$$\pi = \dfrac{1.394 \times 10^{-2}}{4.5 \times 10^{-4}}$$

$$\pi = 30.96 \, \mathrm{Pa}$$

Molar mass of benzoic acid $$\mathrm{C_6H_5COOH} = 122 \, \mathrm{g \, mol^{-1}}$$ (the normal, un-associated value).

Step 1 — Observed molar mass from the freezing point depression.

From $$\Delta T_f = \dfrac{K_f \, w_2 \times 1000}{M_2 \, w_1}$$, the experimentally observed molar mass is:

$$M_{\text{obs}} = \dfrac{K_f \, w_2 \times 1000}{\Delta T_f \, w_1} = \dfrac{4.9 \times 2 \times 1000}{1.62 \times 25} = \dfrac{9800}{40.5}$$

$$M_{\text{obs}} = 242 \, \mathrm{g \, mol^{-1}}$$

Step 2 — van't Hoff factor.

When the solute associates, the van't Hoff factor is the ratio of the normal to the observed molar mass:

$$i = \dfrac{M_{\text{normal}}}{M_{\text{obs}}} = \dfrac{122}{242} = 0.504$$

Step 3 — Degree of association.

Benzoic acid dimerises: $$2\,\mathrm{C_6H_5COOH} \rightleftharpoons (\mathrm{C_6H_5COOH})_2$$. If a fraction $$\alpha$$ of the molecules associate, with $$n = 2$$ molecules forming one dimer:

$$i = 1 - \alpha + \dfrac{\alpha}{n} = 1 - \dfrac{\alpha}{2}$$

Therefore:

$$0.504 = 1 - \dfrac{\alpha}{2} \quad\Rightarrow\quad \dfrac{\alpha}{2} = 0.496 \quad\Rightarrow\quad \alpha = 0.992$$

$$\text{Percentage association} = 99.2\,\%$$

Step 1 — Moles and molality of acetic acid.

Mass of acetic acid $$= \text{volume} \times \text{density} = 0.6 \, \mathrm{mL} \times 1.06 \, \mathrm{g \, mL^{-1}} = 0.636 \, \mathrm{g}$$

Molar mass of $$\mathrm{CH_3COOH} = 60 \, \mathrm{g \, mol^{-1}}$$, so the moles of acid are:

$$n = \dfrac{0.636}{60} = 0.0106 \, \mathrm{mol}$$

1 litre of water has a mass of about 1 kg, so the molality is $$m = 0.0106 \, \mathrm{mol \, kg^{-1}}$$.

Step 2 — van't Hoff factor.

The freezing point depression expected if the acid did not dissociate:

$$\Delta T_{f,\text{calc}} = K_f \, m = 1.86 \times 0.0106 = 0.0197 \, \mathrm{K}$$

The van't Hoff factor is the ratio of the observed depression to the calculated value:

$$i = \dfrac{\Delta T_{f,\text{obs}}}{\Delta T_{f,\text{calc}}} = \dfrac{0.0205}{0.0197} = 1.041$$

Step 3 — Degree of dissociation.

Acetic acid dissociates as $$\mathrm{CH_3COOH} \rightleftharpoons \mathrm{CH_3COO^-} + \mathrm{H^+}$$, producing 2 particles. For dissociation $$i = 1 + \alpha(n - 1) = 1 + \alpha$$ with $$n = 2$$, so:

$$\alpha = i - 1 = 1.041 - 1 = 0.041$$

Step 4 — Dissociation constant.

With initial concentration $$C = 0.0106 \, \mathrm{mol \, L^{-1}}$$:

$$K_a = \dfrac{C\alpha^2}{1 - \alpha} = \dfrac{0.0106 \times (0.041)^2}{1 - 0.041}$$

$$K_a = \dfrac{0.0106 \times 1.681 \times 10^{-3}}{0.959}$$

$$K_a = 1.86 \times 10^{-5}$$

Solution: A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition and properties are uniform throughout. The component present in the largest amount is called the solvent, and the other component(s) are the solute(s).

Depending on the physical state (solid, liquid or gas) of the solute and of the solvent, nine types of solutions are possible. They are conveniently grouped according to the physical state of the solvent.

Gaseous solutions (solvent is a gas):

• Gas in gas — a mixture of oxygen and nitrogen gases.
• Liquid in gas — water vapour (or chloroform vapour) present in air.
• Solid in gas — camphor vapour in nitrogen gas.

Liquid solutions (solvent is a liquid):

• Gas in liquid — oxygen dissolved in water; $$\mathrm{CO_2}$$ dissolved in water (soda water).
• Liquid in liquid — ethanol dissolved in water.
• Solid in liquid — glucose or common salt dissolved in water.

Solid solutions (solvent is a solid):

• Gas in solid — solution of hydrogen in palladium.
• Liquid in solid — amalgam of mercury with sodium.
• Solid in solid — alloys such as brass, or copper dissolved in gold.

A solid solution is one in which the solvent is a solid. When the dissolved solute is a gas, a familiar example is the solution of hydrogen gas in palladium metal.

Palladium can absorb (occlude) very large volumes of hydrogen gas; the hydrogen atoms occupy the interstitial spaces within the metallic lattice, producing a homogeneous solid solution of a gas dissolved in a solid.

Consider 100 g of the solution. Being 68% nitric acid by mass, it contains 68 g of $$\mathrm{HNO_3}$$.

Molar mass of $$\mathrm{HNO_3} = 1 + 14 + 3(16) = 63 \, \mathrm{g \, mol^{-1}}$$

Number of moles of $$\mathrm{HNO_3}$$:

$$n = \dfrac{68}{63} = 1.079 \, \mathrm{mol}$$

Volume of this 100 g of solution, obtained from its density:

$$V = \dfrac{\text{mass}}{\text{density}} = \dfrac{100 \, \mathrm{g}}{1.504 \, \mathrm{g \, mL^{-1}}} = 66.49 \, \mathrm{mL} = 0.06649 \, \mathrm{L}$$

Molarity of the acid:

$$M = \dfrac{n}{V} = \dfrac{1.079 \, \mathrm{mol}}{0.06649 \, \mathrm{L}}$$

$$M = 16.23 \, \mathrm{M}$$

Consider 100 g of the solution. Being 10% (w/w) glucose, it contains 10 g of glucose and $$100 - 10 = 90 \, \mathrm{g}$$ of water.

Molar masses: glucose $$\mathrm{C_6H_{12}O_6} = 180 \, \mathrm{g \, mol^{-1}}$$; water $$= 18 \, \mathrm{g \, mol^{-1}}$$.

Number of moles:

$$n_{\text{glucose}} = \dfrac{10}{180} = 0.0556 \, \mathrm{mol}$$
$$n_{\text{water}} = \dfrac{90}{18} = 5.0 \, \mathrm{mol}$$

Molality (mass of water $$= 90 \, \mathrm{g} = 0.090 \, \mathrm{kg}$$):

$$m = \dfrac{0.0556}{0.090} = 0.617 \, \mathrm{mol \, kg^{-1}}$$

Mole fractions:

$$x_{\text{glucose}} = \dfrac{0.0556}{0.0556 + 5.0} = \dfrac{0.0556}{5.0556} = 0.011$$

$$x_{\text{water}} = 1 - 0.011 = 0.989$$

Molarity — first obtain the volume of 100 g of solution from its density:

$$V = \dfrac{100 \, \mathrm{g}}{1.2 \, \mathrm{g \, mL^{-1}}} = 83.33 \, \mathrm{mL} = 0.08333 \, \mathrm{L}$$

$$M = \dfrac{n_{\text{glucose}}}{V} = \dfrac{0.0556}{0.08333} = 0.67 \, \mathrm{M}$$

Since the mixture contains equimolar amounts, let the number of moles of $$\mathrm{Na_2CO_3}$$ and of $$\mathrm{NaHCO_3}$$ each be $$x$$.

Molar masses: $$\mathrm{Na_2CO_3} = 2(23) + 12 + 3(16) = 106 \, \mathrm{g \, mol^{-1}}$$; $$\mathrm{NaHCO_3} = 23 + 1 + 12 + 3(16) = 84 \, \mathrm{g \, mol^{-1}}$$.

The total mass of the mixture is 1 g:

$$106x + 84x = 1 \quad\Rightarrow\quad 190x = 1 \quad\Rightarrow\quad x = 5.263 \times 10^{-3} \, \mathrm{mol}$$

The neutralisation reactions with HCl are:

$$\mathrm{Na_2CO_3 + 2HCl \longrightarrow 2NaCl + H_2O + CO_2}$$

$$\mathrm{NaHCO_3 + HCl \longrightarrow NaCl + H_2O + CO_2}$$

So 1 mol of $$\mathrm{Na_2CO_3}$$ needs 2 mol of HCl, and 1 mol of $$\mathrm{NaHCO_3}$$ needs 1 mol of HCl. The total moles of HCl required are:

$$n_{\mathrm{HCl}} = 2x + x = 3x = 3 \times 5.263 \times 10^{-3} = 0.01579 \, \mathrm{mol}$$

Volume of 0.1 M HCl needed:

$$V = \dfrac{n_{\mathrm{HCl}}}{M} = \dfrac{0.01579 \, \mathrm{mol}}{0.1 \, \mathrm{mol \, L^{-1}}} = 0.1579 \, \mathrm{L}$$

$$V = 157.9 \, \mathrm{mL}$$

First find the mass of solute contributed by each of the two solutions.

From the first solution (300 g, 25% by mass):

$$w_1 = \dfrac{25}{100} \times 300 = 75 \, \mathrm{g}$$

From the second solution (400 g, 40% by mass):

$$w_2 = \dfrac{40}{100} \times 400 = 160 \, \mathrm{g}$$

Total mass of solute in the mixed solution:

$$w = 75 + 160 = 235 \, \mathrm{g}$$

Total mass of the resulting solution:

$$m = 300 + 400 = 700 \, \mathrm{g}$$

Mass percentage of solute:

$$\% \text{ solute} = \dfrac{235}{700} \times 100 = 33.57\,\%$$

Mass percentage of the solvent (water):

$$\% \text{ water} = 100 - 33.57 = 66.43\,\%$$

Molar mass of ethylene glycol $$\mathrm{C_2H_6O_2} = 2(12) + 6(1) + 2(16) = 62 \, \mathrm{g \, mol^{-1}}$$.

Number of moles of ethylene glycol:

$$n = \dfrac{222.6}{62} = 3.59 \, \mathrm{mol}$$

Molality — the mass of water (solvent) is $$200 \, \mathrm{g} = 0.200 \, \mathrm{kg}$$:

$$m = \dfrac{3.59 \, \mathrm{mol}}{0.200 \, \mathrm{kg}} = 17.95 \, \mathrm{mol \, kg^{-1}}$$

Molarity — total mass of the solution:

$$m_{\text{soln}} = 222.6 + 200 = 422.6 \, \mathrm{g}$$

Volume of the solution from its density:

$$V = \dfrac{422.6 \, \mathrm{g}}{1.072 \, \mathrm{g \, mL^{-1}}} = 394.2 \, \mathrm{mL} = 0.3942 \, \mathrm{L}$$

$$M = \dfrac{3.59 \, \mathrm{mol}}{0.3942 \, \mathrm{L}} = 9.11 \, \mathrm{M}$$

In pure ethanol (alcohol) the molecules are held together by hydrogen bonds, and in pure water the molecules are also strongly hydrogen-bonded to one another.

When alcohol and water are mixed, the new hydrogen bonds formed between the alcohol and water molecules are weaker than the average strength of the alcohol–alcohol and water–water hydrogen bonds that existed in the pure liquids. In effect, the overall intermolecular attractive forces become weaker on mixing.

Because the molecules are now held less tightly, they escape into the vapour phase more readily. Consequently the vapour pressure of the solution is greater than that predicted by Raoult's law — the solution shows a positive deviation. The mixing is also accompanied by absorption of heat ($$\Delta_{\mathrm{mix}} H > 0$$) and a small increase in volume ($$\Delta_{\mathrm{mix}} V > 0$$).

When a gas dissolves in a liquid, gas molecules pass from the gaseous state into the solution. This can be represented as an equilibrium:

$$\text{Gas} + \text{Liquid} \rightleftharpoons \text{Gas dissolved in liquid}$$

The dissolution of a gas in a liquid is an exothermic process ($$\Delta H < 0$$), since the gas molecules become attached to the solvent molecules with release of energy.

By Le Chatelier's principle, increasing the temperature shifts an exothermic equilibrium in the backward direction — that is, towards the expulsion of dissolved gas. Hence the solubility of the gas decreases as the temperature rises.

Moreover, at a higher temperature the dissolved gas molecules have greater kinetic energy and can more easily overcome the attractive forces of the solvent and escape from the solution.

Henry's law: At a constant temperature, the partial pressure of a gas in the vapour phase ($$p$$) is directly proportional to the mole fraction of the gas ($$x$$) dissolved in the solution.

$$p = K_H \, x$$

where $$K_H$$ is the Henry's law constant. A larger value of $$K_H$$ corresponds to a lower solubility of the gas.

Important applications:

• To increase the solubility of $$\mathrm{CO_2}$$ in soft drinks and soda water, the bottles are sealed under a high pressure of carbon dioxide.
• Scuba divers breathe compressed air under water, which raises the solubility of atmospheric gases in the blood. To prevent the painful condition called bends (bubbles of nitrogen forming in the blood during decompression), the cylinders used by divers are filled with air diluted with helium (roughly 11.7% He, 56.2% $$\mathrm{N_2}$$, 32.1% $$\mathrm{O_2}$$).
• At high altitudes the partial pressure of oxygen is low, which lowers the concentration of oxygen dissolved in the blood. This causes anoxia, in which climbers feel weak and are unable to think clearly.

By Henry's law the amount of a gas dissolved in a fixed quantity of solvent is directly proportional to the partial pressure of the gas above the solution.

Hence, for the same solvent, the ratio of partial pressure to mass of dissolved gas is constant:

$$\dfrac{p_1}{m_1} = \dfrac{p_2}{m_2}$$

Given $$p_1 = 1 \, \mathrm{bar}$$ for $$m_1 = 6.56 \times 10^{-3} \, \mathrm{g}$$ of ethane, and $$m_2 = 5.00 \times 10^{-2} \, \mathrm{g}$$:

$$p_2 = p_1 \times \dfrac{m_2}{m_1} = 1 \times \dfrac{5.00 \times 10^{-2}}{6.56 \times 10^{-3}}$$

$$p_2 = 7.62 \, \mathrm{bar}$$

An ideal solution obeys Raoult's law over the entire composition range, and for it $$\Delta_{\mathrm{mix}} H = 0$$ and $$\Delta_{\mathrm{mix}} V = 0$$. Real solutions often deviate from this behaviour.

Positive deviation from Raoult's law: The observed partial vapour pressure of each component, and hence the total vapour pressure, is higher than predicted by Raoult's law. This occurs when the solute–solvent (A–B) interactions are weaker than the solute–solute (A–A) and solvent–solvent (B–B) interactions. Replacing stronger A–A and B–B attractions by weaker A–B attractions absorbs energy, so the mixing is endothermic: $$\Delta_{\mathrm{mix}} H > 0$$ (also $$\Delta_{\mathrm{mix}} V > 0$$). Example: a mixture of ethanol and acetone.

Negative deviation from Raoult's law: The observed vapour pressure is lower than predicted by Raoult's law. This occurs when the A–B interactions are stronger than the A–A and B–B interactions. Formation of these stronger attractions releases energy, so the mixing is exothermic: $$\Delta_{\mathrm{mix}} H < 0$$ (also $$\Delta_{\mathrm{mix}} V < 0$$). Example: a mixture of chloroform and acetone.

Thus a positive deviation is associated with $$\Delta_{\mathrm{mix}} H > 0$$, and a negative deviation with $$\Delta_{\mathrm{mix}} H < 0$$.

At the normal boiling point of the pure solvent (water), the vapour pressure of pure water equals the normal atmospheric pressure, $$p^{\circ} = 1.013 \, \mathrm{bar}$$.

The vapour pressure of the solution at the same temperature is $$p = 1.004 \, \mathrm{bar}$$.

A 2% solution contains 2 g of solute in 100 g of solution, i.e. 2 g of solute and 98 g of water.

By Raoult's law, the relative lowering of vapour pressure equals the mole fraction of the solute. For a dilute solution:

$$\dfrac{p^{\circ} - p}{p^{\circ}} = \dfrac{n_2}{n_1 + n_2} \approx \dfrac{n_2}{n_1} = \dfrac{w_2 / M_2}{w_1 / M_1}$$

Number of moles of water:

$$n_1 = \dfrac{98}{18} = 5.444 \, \mathrm{mol}$$

Relative lowering of vapour pressure:

$$\dfrac{p^{\circ} - p}{p^{\circ}} = \dfrac{1.013 - 1.004}{1.013} = \dfrac{0.009}{1.013} = 0.00888$$

Therefore:

$$0.00888 = \dfrac{2 / M_2}{5.444}$$

$$\dfrac{2}{M_2} = 0.00888 \times 5.444 = 0.04834$$

$$M_2 = \dfrac{2}{0.04834}$$

$$M_2 = 41.35 \, \mathrm{g \, mol^{-1}}$$

Molar masses: heptane $$\mathrm{C_7H_{16}} = 7(12) + 16(1) = 100 \, \mathrm{g \, mol^{-1}}$$; octane $$\mathrm{C_8H_{18}} = 8(12) + 18(1) = 114 \, \mathrm{g \, mol^{-1}}$$.

Number of moles of each component:

$$n_{\text{heptane}} = \dfrac{26.0}{100} = 0.260 \, \mathrm{mol}$$
$$n_{\text{octane}} = \dfrac{35.0}{114} = 0.307 \, \mathrm{mol}$$

Total moles $$= 0.260 + 0.307 = 0.567 \, \mathrm{mol}$$.

Mole fractions in the liquid:

$$x_{\text{heptane}} = \dfrac{0.260}{0.567} = 0.4585$$

$$x_{\text{octane}} = \dfrac{0.307}{0.567} = 0.5415$$

For an ideal solution, the total vapour pressure is given by Raoult's law:

$$p_{\text{total}} = p^{\circ}_{\text{heptane}} \, x_{\text{heptane}} + p^{\circ}_{\text{octane}} \, x_{\text{octane}}$$

$$p_{\text{total}} = (105.2)(0.4585) + (46.8)(0.5415)$$

$$p_{\text{total}} = 48.24 + 25.34$$

$$p_{\text{total}} = 73.58 \, \mathrm{kPa}$$

A 1 molal solution contains 1 mol of the non-volatile solute dissolved in 1 kg (1000 g) of water.

Number of moles of water:

$$n_1 = \dfrac{1000}{18} = 55.56 \, \mathrm{mol}$$

Number of moles of solute: $$n_2 = 1 \, \mathrm{mol}$$.

Mole fraction of the solute:

$$x_2 = \dfrac{n_2}{n_1 + n_2} = \dfrac{1}{1 + 55.56} = \dfrac{1}{56.56} = 0.0177$$

By Raoult's law, the relative lowering of vapour pressure equals the mole fraction of the solute:

$$\dfrac{p^{\circ} - p}{p^{\circ}} = x_2$$

So the lowering of vapour pressure is:

$$p^{\circ} - p = x_2 \, p^{\circ} = 0.0177 \times 12.3 = 0.218 \, \mathrm{kPa}$$

Vapour pressure of the solution:

$$p = 12.3 - 0.218 = 12.08 \, \mathrm{kPa}$$

Let the vapour pressure of pure octane be $$p^{\circ}$$. The vapour pressure of the solution is reduced to 80% of this, so $$p = 0.80 \, p^{\circ}$$.

Relative lowering of vapour pressure:

$$\dfrac{p^{\circ} - p}{p^{\circ}} = \dfrac{p^{\circ} - 0.80 \, p^{\circ}}{p^{\circ}} = 0.20$$

By Raoult's law, the relative lowering of vapour pressure equals the mole fraction of the solute. For a dilute solution:

$$\dfrac{p^{\circ} - p}{p^{\circ}} = \dfrac{n_2}{n_1} = \dfrac{w_2 / M_2}{w_1 / M_1}$$

Molar mass of octane $$\mathrm{C_8H_{18}} = 114 \, \mathrm{g \, mol^{-1}}$$, so the moles of octane are:

$$n_1 = \dfrac{114}{114} = 1 \, \mathrm{mol}$$

Substituting (solute molar mass $$M_2 = 40 \, \mathrm{g \, mol^{-1}}$$):

$$0.20 = \dfrac{w_2 / 40}{1}$$

$$w_2 = 0.20 \times 40$$

$$w_2 = 8 \, \mathrm{g}$$

Step 1 — Find $$K_f$$ of water using the cane sugar solution.

A 5% (by mass) solution contains 5 g of sugar in 100 g of solution, i.e. 5 g of sugar in 95 g of water.

Molar mass of cane sugar (sucrose) $$\mathrm{C_{12}H_{22}O_{11}} = 342 \, \mathrm{g \, mol^{-1}}$$.

Molality of the sugar solution:

$$m_{\text{sugar}} = \dfrac{5/342}{95/1000} = \dfrac{0.01462}{0.095} = 0.1539 \, \mathrm{mol \, kg^{-1}}$$

Depression observed for the sugar solution:

$$\Delta T_f = 273.15 - 271 = 2.15 \, \mathrm{K}$$

From $$\Delta T_f = K_f \, m$$:

$$K_f = \dfrac{\Delta T_f}{m_{\text{sugar}}} = \dfrac{2.15}{0.1539} = 13.97 \, \mathrm{K \, kg \, mol^{-1}}$$

Step 2 — Apply $$K_f$$ to the 5% glucose solution.

A 5% glucose solution likewise has 5 g of glucose in 95 g of water. Molar mass of glucose $$\mathrm{C_6H_{12}O_6} = 180 \, \mathrm{g \, mol^{-1}}$$.

$$m_{\text{glucose}} = \dfrac{5/180}{95/1000} = \dfrac{0.02778}{0.095} = 0.2924 \, \mathrm{mol \, kg^{-1}}$$

Depression of freezing point for the glucose solution:

$$\Delta T_f = K_f \, m_{\text{glucose}} = 13.97 \times 0.2924 = 4.09 \, \mathrm{K}$$

Freezing point of the glucose solution:

$$T_f = 273.15 - 4.09 = 269.06 \, \mathrm{K}$$

The molar mass of a solute is related to the freezing point depression by:

$$M_2 = \dfrac{K_f \, w_2 \times 1000}{\Delta T_f \, w_1}$$

where $$w_1$$ is the mass of the solvent (benzene) in grams.

For $$\mathrm{AB_2}$$: $$w_2 = 1 \, \mathrm{g}$$, $$w_1 = 20 \, \mathrm{g}$$, $$\Delta T_f = 2.3 \, \mathrm{K}$$.

$$M_{\mathrm{AB_2}} = \dfrac{5.1 \times 1 \times 1000}{2.3 \times 20} = \dfrac{5100}{46} = 110.87 \, \mathrm{g \, mol^{-1}}$$

For $$\mathrm{AB_4}$$: $$w_2 = 1.0 \, \mathrm{g}$$, $$w_1 = 20 \, \mathrm{g}$$, $$\Delta T_f = 1.3 \, \mathrm{K}$$.

$$M_{\mathrm{AB_4}} = \dfrac{5.1 \times 1 \times 1000}{1.3 \times 20} = \dfrac{5100}{26} = 196.15 \, \mathrm{g \, mol^{-1}}$$

Let the atomic masses of A and B be $$a$$ and $$b$$ respectively. From the formulae:

$$a + 2b = 110.87 \qquad (1)$$

$$a + 4b = 196.15 \qquad (2)$$

Subtracting (1) from (2):

$$2b = 196.15 - 110.87 = 85.28 \quad\Rightarrow\quad b = 42.64$$

From (1):

$$a = 110.87 - 2(42.64) = 110.87 - 85.28 = 25.59$$

$$\text{Atomic mass of A} = 25.59, \qquad \text{atomic mass of B} = 42.64$$

For the first solution, the molar concentration of glucose is:

$$C_1 = \dfrac{\text{moles of glucose}}{\text{volume in litres}} = \dfrac{36/180}{1} = 0.2 \, \mathrm{mol \, L^{-1}}$$

(Molar mass of glucose $$= 180 \, \mathrm{g \, mol^{-1}}$$.)

The osmotic pressure obeys $$\pi = C R T$$. At constant temperature, $$\dfrac{\pi}{C} = R T$$ is constant, so for the two solutions:

$$\dfrac{\pi_1}{C_1} = \dfrac{\pi_2}{C_2}$$

Solving for the required concentration:

$$C_2 = C_1 \times \dfrac{\pi_2}{\pi_1} = 0.2 \times \dfrac{1.52}{4.98}$$

$$C_2 = 0.061 \, \mathrm{mol \, L^{-1}}$$

n-Octane is a non-polar solvent. According to the principle like dissolves like, non-polar solutes dissolve readily in it, whereas polar and ionic solutes dissolve poorly.

Examining each substance:

• Cyclohexane — a non-polar hydrocarbon, structurally very similar to octane; therefore freely (most) soluble.
• $$\mathrm{CH_3CN}$$ (acetonitrile) — a polar molecule, but only moderately polar; partly soluble.
• $$\mathrm{CH_3OH}$$ (methanol) — polar and capable of strong hydrogen bonding; less soluble than acetonitrile.
• KCl — an ionic compound with very strong ion–ion forces; non-polar octane cannot solvate the ions, so it is the least soluble.

Hence the order of increasing solubility in n-octane is:

$$\mathrm{KCl} < \mathrm{CH_3OH} < \mathrm{CH_3CN} < \text{cyclohexane}$$

Consider a sample of the lake water containing 1 kg (1000 g) of water. It contains 92 g of $$\mathrm{Na^+}$$ ions.

Number of moles of $$\mathrm{Na^+}$$ (atomic mass of Na $$= 23$$):

$$n_{\mathrm{Na^+}} = \dfrac{92}{23} = 4 \, \mathrm{mol}$$

Total mass of this solution:

$$m_{\text{soln}} = 1000 + 92 = 1092 \, \mathrm{g}$$

Volume of the solution, obtained from its density:

$$V = \dfrac{m_{\text{soln}}}{\text{density}} = \dfrac{1092 \, \mathrm{g}}{1.25 \, \mathrm{g \, mL^{-1}}} = 873.6 \, \mathrm{mL} = 0.8736 \, \mathrm{L}$$

Molarity of $$\mathrm{Na^+}$$ ions:

$$M = \dfrac{n_{\mathrm{Na^+}}}{V} = \dfrac{4}{0.8736}$$

$$M = 4.58 \, \mathrm{M}$$

CuS is a sparingly soluble salt. It dissolves to a small extent according to:

$$\mathrm{CuS\;(s) \rightleftharpoons Cu^{2+}\;(aq) + S^{2-}\;(aq)}$$

Let the maximum solubility (molarity) of CuS be $$s \, \mathrm{mol \, L^{-1}}$$. At saturation, each formula unit that dissolves gives one $$\mathrm{Cu^{2+}}$$ and one $$\mathrm{S^{2-}}$$ ion, so $$[\mathrm{Cu^{2+}}] = s$$ and $$[\mathrm{S^{2-}}] = s$$.

The solubility product is:

$$K_{sp} = [\mathrm{Cu^{2+}}][\mathrm{S^{2-}}] = s \times s = s^2$$

Therefore the maximum molarity is:

$$s = \sqrt{K_{sp}} = \sqrt{6 \times 10^{-16}}$$

$$s = 2.45 \times 10^{-8} \, \mathrm{mol \, L^{-1}}$$

The mass percentage of a component is its mass divided by the total mass of the solution, multiplied by 100.

Total mass of the solution:

$$m_{\text{total}} = 6.5 + 450 = 456.5 \, \mathrm{g}$$

Mass percentage of aspirin:

$$\%\,\text{aspirin} = \dfrac{6.5}{456.5} \times 100$$

$$\%\,\text{aspirin} = 1.424\,\%$$

Molar mass of nalorphene $$\mathrm{C_{19}H_{21}NO_3}$$:

$$= 19(12) + 21(1) + 14 + 3(16)$$
$$= 228 + 21 + 14 + 48 = 311 \, \mathrm{g \, mol^{-1}}$$

A single dose contains 1.5 mg of nalorphene. Number of moles in the dose:

$$n = \dfrac{1.5 \times 10^{-3} \, \mathrm{g}}{311 \, \mathrm{g \, mol^{-1}}} = 4.823 \times 10^{-6} \, \mathrm{mol}$$

The solution is $$1.5 \times 10^{-3}$$ molal, i.e. it contains $$1.5 \times 10^{-3} \, \mathrm{mol}$$ of nalorphene per kilogram of water. The mass of water that holds the required moles is:

$$\text{mass of water} = \dfrac{n}{\text{molality}} = \dfrac{4.823 \times 10^{-6}}{1.5 \times 10^{-3}} = 3.215 \times 10^{-3} \, \mathrm{kg} = 3.215 \, \mathrm{g}$$

Mass of the solution = mass of water + mass of nalorphene:

$$m_{\text{soln}} = 3.215 + 0.0015 = 3.217 \, \mathrm{g}$$

$$m_{\text{soln}} \approx 3.22 \, \mathrm{g}$$

Number of moles of benzoic acid required for the solution:

$$n = M \times V = 0.15 \, \mathrm{mol \, L^{-1}} \times 0.250 \, \mathrm{L} = 0.0375 \, \mathrm{mol}$$

Molar mass of benzoic acid $$\mathrm{C_6H_5COOH} = 7(12) + 6(1) + 2(16) = 122 \, \mathrm{g \, mol^{-1}}$$

Mass of benzoic acid required:

$$w = n \times M = 0.0375 \times 122$$

$$w = 4.575 \, \mathrm{g}$$

The depression in freezing point is a colligative property — it depends on the number of solute particles present in solution. The more an acid ionises, the more $$\mathrm{H^+}$$ and anion particles it produces, and the larger the depression.

All three are carboxylic acids that ionise to give hydrogen ions; the extent of ionisation depends on the acid strength.

In acetic acid ($$\mathrm{CH_3COOH}$$) the methyl group is electron-releasing and slightly destabilises the carboxylate ion, so it is the weakest acid and ionises the least.

In trichloroacetic acid ($$\mathrm{CCl_3COOH}$$) the three chlorine atoms are strongly electron-withdrawing (–I effect). They draw electron density away from the –COOH group, stabilising the carboxylate ion and making the acid much stronger, so it ionises to a greater extent.

In trifluoroacetic acid ($$\mathrm{CF_3COOH}$$) fluorine is even more electronegative than chlorine, so its –I effect is the strongest. This makes trifluoroacetic acid the strongest of the three, and it ionises the most.

Since the degree of ionisation increases in the order acetic acid < trichloroacetic acid < trifluoroacetic acid, the number of particles in solution — and therefore the depression in freezing point — increases in exactly the same order.

The compound is 2-chlorobutanoic acid, $$\mathrm{CH_3CH_2CHClCOOH}$$, with molecular formula $$\mathrm{C_4H_7ClO_2}$$.

Molar mass $$= 4(12) + 7(1) + 35.5 + 2(16) = 48 + 7 + 35.5 + 32 = 122.5 \, \mathrm{g \, mol^{-1}}$$

Number of moles of the acid:

$$n = \dfrac{10}{122.5} = 0.0816 \, \mathrm{mol}$$

Molality (mass of water $$= 250 \, \mathrm{g} = 0.250 \, \mathrm{kg}$$):

$$m = \dfrac{0.0816}{0.250} = 0.3265 \, \mathrm{mol \, kg^{-1}}$$

The acid dissociates as $$\mathrm{HA \rightleftharpoons H^+ + A^-}$$. For a weak acid of concentration $$C$$ and degree of dissociation $$\alpha$$, $$K_a = \dfrac{C\alpha^2}{1-\alpha} \approx C\alpha^2$$. Hence:

$$\alpha = \sqrt{\dfrac{K_a}{C}} = \sqrt{\dfrac{1.4 \times 10^{-3}}{0.3265}} = \sqrt{4.288 \times 10^{-3}} = 0.0655$$

Each molecule that dissociates produces 2 particles, so the van't Hoff factor is:

$$i = 1 + \alpha = 1 + 0.0655 = 1.0655$$

The depression in freezing point of an electrolyte solution is $$\Delta T_f = i \, K_f \, m$$:

$$\Delta T_f = 1.0655 \times 1.86 \times 0.3265$$

$$\Delta T_f \approx 0.65 \, \mathrm{K}$$

Molar mass of fluoroacetic acid $$\mathrm{CH_2FCOOH}$$ (formula $$\mathrm{C_2H_3FO_2}$$):

$$= 2(12) + 3(1) + 19 + 2(16) = 24 + 3 + 19 + 32 = 78 \, \mathrm{g \, mol^{-1}}$$

Number of moles of the acid:

$$n = \dfrac{19.5}{78} = 0.25 \, \mathrm{mol}$$

Molality (mass of water $$= 500 \, \mathrm{g} = 0.500 \, \mathrm{kg}$$):

$$m = \dfrac{0.25}{0.500} = 0.5 \, \mathrm{mol \, kg^{-1}}$$

van't Hoff factor. The depression of freezing point expected if the acid did not dissociate:

$$\Delta T_{f,\text{calc}} = K_f \, m = 1.86 \times 0.5 = 0.93 \, \mathrm{K}$$

The van't Hoff factor is the ratio of the observed depression to the calculated value:

$$i = \dfrac{\Delta T_{f,\text{obs}}}{\Delta T_{f,\text{calc}}} = \dfrac{1.0}{0.93} = 1.0753$$

Dissociation constant. The acid dissociates as $$\mathrm{CH_2FCOOH \rightleftharpoons CH_2FCOO^- + H^+}$$, producing 2 particles, so $$i = 1 + \alpha$$:

$$\alpha = i - 1 = 1.0753 - 1 = 0.0753$$

For the dissociation equilibrium with initial concentration $$C = 0.5 \, \mathrm{mol \, L^{-1}}$$:

$$K_a = \dfrac{C\alpha^2}{1 - \alpha} = \dfrac{0.5 \times (0.0753)^2}{1 - 0.0753}$$

$$K_a = \dfrac{0.5 \times 5.670 \times 10^{-3}}{0.9247}$$

$$K_a = 3.07 \times 10^{-3}$$

Molar mass of glucose $$\mathrm{C_6H_{12}O_6} = 180 \, \mathrm{g \, mol^{-1}}$$ and molar mass of water $$= 18 \, \mathrm{g \, mol^{-1}}$$.

Number of moles:

$$n_{\text{glucose}} = \dfrac{25}{180} = 0.1389 \, \mathrm{mol}$$
$$n_{\text{water}} = \dfrac{450}{18} = 25 \, \mathrm{mol}$$

By Raoult's law, the relative lowering of vapour pressure equals the mole fraction of the solute:

$$\dfrac{p^{\circ} - p}{p^{\circ}} = x_{\text{glucose}} = \dfrac{n_{\text{glucose}}}{n_{\text{glucose}} + n_{\text{water}}}$$

$$\dfrac{p^{\circ} - p}{p^{\circ}} = \dfrac{0.1389}{0.1389 + 25} = \dfrac{0.1389}{25.139} = 0.005525$$

Lowering of vapour pressure:

$$p^{\circ} - p = 0.005525 \times 17.535 = 0.0969 \, \mathrm{mm \, Hg}$$

Vapour pressure of water above the solution:

$$p = 17.535 - 0.0969 = 17.44 \, \mathrm{mm \, Hg}$$

Here the Henry's law constant is given on the molality scale, so Henry's law takes the form:

$$p = K_H \, m$$

where $$m$$ is the molality (the solubility) of the dissolved gas.

Solving for the solubility:

$$m = \dfrac{p}{K_H} = \dfrac{760 \, \mathrm{mm \, Hg}}{4.27 \times 10^5 \, \mathrm{mm \, Hg}}$$

$$m = 1.78 \times 10^{-3} \, \mathrm{mol \, kg^{-1}}$$

Number of moles of each liquid:

$$n_A = \dfrac{100}{140} = 0.714 \, \mathrm{mol}$$
$$n_B = \dfrac{1000}{180} = 5.556 \, \mathrm{mol}$$

Total moles $$= 0.714 + 5.556 = 6.270 \, \mathrm{mol}$$.

Mole fractions:

$$x_A = \dfrac{0.714}{6.270} = 0.1139, \qquad x_B = \dfrac{5.556}{6.270} = 0.8861$$

For an ideal solution the total vapour pressure is the sum of the partial pressures:

$$p_{\text{total}} = p^{\circ}_A \, x_A + p^{\circ}_B \, x_B$$

The partial pressure of B is:

$$p_B = p^{\circ}_B \, x_B = 500 \times 0.8861 = 443.0 \, \mathrm{torr}$$

Hence the partial pressure (vapour pressure in the solution) of A is:

$$p_A = p_{\text{total}} - p_B = 475 - 443.0 = 32.0 \, \mathrm{torr}$$

The vapour pressure of pure A follows from $$p_A = p^{\circ}_A \, x_A$$:

$$p^{\circ}_A = \dfrac{p_A}{x_A} = \dfrac{32.0}{0.1139}$$

$$p^{\circ}_A \approx 280.7 \, \mathrm{torr}, \qquad p_A \approx 32.0 \, \mathrm{torr}$$

Ideal-solution (Raoult's law) lines. For an ideal solution the partial pressures are straight lines in $$x_{\text{acetone}}$$:

$$p_{\text{acetone}} = p^{\circ}_{\text{acetone}} \, x_{\text{acetone}} = 741.8 \, x_{\text{acetone}}$$

$$p_{\text{chloroform}} = p^{\circ}_{\text{chloroform}} \,(1 - x_{\text{acetone}}) = 632.8 \,(1 - x_{\text{acetone}})$$

$$p_{\text{total}} = p_{\text{acetone}} + p_{\text{chloroform}}$$

On the graph (vapour pressure on the y-axis, $$x_{\text{acetone}}$$ on the x-axis), the ideal lines are:

• $$p_{\text{acetone}}$$ — a straight line rising from 0 (at $$x_{\text{acetone}} = 0$$) to 741.8 mm Hg (at $$x_{\text{acetone}} = 1$$).
• $$p_{\text{chloroform}}$$ — a straight line falling from 632.8 mm Hg (at $$x_{\text{acetone}} = 0$$) to 0 (at $$x_{\text{acetone}} = 1$$).
• $$p_{\text{total}}$$ — a straight line joining 632.8 mm Hg to 741.8 mm Hg.

Experimental data. Adding the two experimental partial pressures at each composition gives the experimental total pressure, which can be compared with the ideal value:

When the experimental points for $$p_{\text{acetone}}$$, $$p_{\text{chloroform}}$$ and $$p_{\text{total}}$$ are plotted on the same graph as the ideal straight lines, every experimental curve lies below the corresponding ideal line, and the experimental total pressure passes through a minimum (near $$x_{\text{acetone}} \approx 0.36$$).

Because the observed vapour pressures are lower than those predicted by Raoult's law, the acetone–chloroform mixture shows a negative deviation from ideal behaviour. The cause is the strong attraction between acetone and chloroform molecules through hydrogen bonding (the chloroform C–H bonds to the acetone carbonyl oxygen), which makes the A–B interactions stronger than the A–A and B–B interactions.

Molar masses: benzene $$\mathrm{C_6H_6} = 78 \, \mathrm{g \, mol^{-1}}$$; toluene $$\mathrm{C_7H_8} = 92 \, \mathrm{g \, mol^{-1}}$$.

Number of moles:

$$n_{\text{benzene}} = \dfrac{80}{78} = 1.026 \, \mathrm{mol}$$
$$n_{\text{toluene}} = \dfrac{100}{92} = 1.087 \, \mathrm{mol}$$

Total moles $$= 1.026 + 1.087 = 2.113 \, \mathrm{mol}$$.

Mole fractions in the liquid:

$$x_{\text{benzene}} = \dfrac{1.026}{2.113} = 0.486, \qquad x_{\text{toluene}} = 0.514$$

Partial pressures by Raoult's law:

$$p_{\text{benzene}} = p^{\circ}_{\text{benzene}} \, x_{\text{benzene}} = 50.71 \times 0.486 = 24.65 \, \mathrm{mm \, Hg}$$

$$p_{\text{toluene}} = p^{\circ}_{\text{toluene}} \, x_{\text{toluene}} = 32.06 \times 0.514 = 16.48 \, \mathrm{mm \, Hg}$$

Total vapour pressure:

$$p_{\text{total}} = 24.65 + 16.48 = 41.13 \, \mathrm{mm \, Hg}$$

The mole fraction of benzene in the vapour phase is its partial pressure divided by the total pressure:

$$y_{\text{benzene}} = \dfrac{p_{\text{benzene}}}{p_{\text{total}}} = \dfrac{24.65}{41.13}$$

$$y_{\text{benzene}} = 0.60$$

The total pressure of the air over the water is 10 atm. Converting to mm Hg:

$$P = 10 \, \mathrm{atm} = 10 \times 760 = 7600 \, \mathrm{mm \, Hg}$$

The partial pressure of each gas equals its volume fraction times the total pressure:

$$p_{\mathrm{O_2}} = \dfrac{20}{100} \times 7600 = 1520 \, \mathrm{mm \, Hg}$$

$$p_{\mathrm{N_2}} = \dfrac{79}{100} \times 7600 = 6004 \, \mathrm{mm \, Hg}$$

By Henry's law, the mole fraction of each dissolved gas is $$x = \dfrac{p}{K_H}$$.

For oxygen:

$$x_{\mathrm{O_2}} = \dfrac{p_{\mathrm{O_2}}}{K_H(\mathrm{O_2})} = \dfrac{1520}{3.30 \times 10^7} = 4.61 \times 10^{-5}$$

For nitrogen:

$$x_{\mathrm{N_2}} = \dfrac{p_{\mathrm{N_2}}}{K_H(\mathrm{N_2})} = \dfrac{6004}{6.51 \times 10^7} = 9.22 \times 10^{-5}$$

For an electrolyte, the osmotic pressure is $$\pi = i \, C R T = i \, \dfrac{n}{V} R T$$, where $$i$$ is the van't Hoff factor.

Solving for the number of moles of $$\mathrm{CaCl_2}$$:

$$n = \dfrac{\pi \, V}{i \, R T}$$

Data: $$\pi = 0.75 \, \mathrm{atm}$$, $$V = 2.5 \, \mathrm{L}$$, $$i = 2.47$$, $$R = 0.0821 \, \mathrm{L \, atm \, K^{-1} \, mol^{-1}}$$ and $$T = 27\ ^{\circ}\mathrm{C} = 300 \, \mathrm{K}$$.

$$n = \dfrac{0.75 \times 2.5}{2.47 \times 0.0821 \times 300} = \dfrac{1.875}{60.84}$$

$$n = 0.0308 \, \mathrm{mol}$$

Molar mass of $$\mathrm{CaCl_2} = 40 + 2(35.5) = 111 \, \mathrm{g \, mol^{-1}}$$

Mass of $$\mathrm{CaCl_2}$$ to be dissolved:

$$w = n \times M = 0.0308 \times 111$$

$$w \approx 3.42 \, \mathrm{g}$$

$$\mathrm{K_2SO_4}$$ dissociates completely as $$\mathrm{K_2SO_4 \longrightarrow 2K^+ + SO_4^{2-}}$$, giving 3 ions per formula unit. Hence the van't Hoff factor is $$i = 3$$.

Molar mass of $$\mathrm{K_2SO_4} = 2(39) + 32 + 4(16) = 78 + 32 + 64 = 174 \, \mathrm{g \, mol^{-1}}$$

Number of moles ($$25 \, \mathrm{mg} = 25 \times 10^{-3} \, \mathrm{g}$$):

$$n = \dfrac{25 \times 10^{-3}}{174} = 1.437 \times 10^{-4} \, \mathrm{mol}$$

The osmotic pressure of an electrolyte solution is:

$$\pi = i \, \dfrac{n}{V} R T$$

Data: $$i = 3$$, $$V = 2 \, \mathrm{L}$$, $$R = 0.0821 \, \mathrm{L \, atm \, K^{-1} \, mol^{-1}}$$, $$T = 25\ ^{\circ}\mathrm{C} = 298 \, \mathrm{K}$$.

$$\pi = \dfrac{3 \times (1.437 \times 10^{-4}) \times 0.0821 \times 298}{2}$$

$$\pi = \dfrac{1.055 \times 10^{-2}}{2}$$

$$\pi = 5.27 \times 10^{-3} \, \mathrm{atm}$$