JEE Chemistry Solutions Notes Set 02

Expressing Concentration

A solution is a homogeneous mixture; the component present in larger amount is the solvent. The common concentration terms are:
Mass percentage = (mass of solute/mass of solution) × 100; ppm = (mass of solute/mass of solution) × 10⁶.
Mole fraction of a component, \( x_i=\frac{n_i}{\sum n} \), with the sum of all mole fractions equal to 1. It does not change with temperature.
Molarity, \( M=\frac{n_{solute}}{V_{solution}(L)} \), changes with temperature because volume changes. Molality, \( m=\frac{n_{solute}}{\text{mass of solvent (kg)}} \), is temperature independent. Normality = molarity × n-factor.
Useful conversion: \( M=\frac{10\times(\text{mass \%})\times d}{\text{molar mass}} \) with d in g cm⁻³. When solutions of the same solute are mixed, add moles and volumes: \( M=\frac{M_1V_1+M_2V_2}{V_1+V_2} \).

Solubility and Henry's Law

The solubility of a solid in a liquid rises with temperature if dissolution is endothermic. For a gas dissolving in a liquid, solubility falls as temperature rises and rises with pressure.
Henry's law: the partial pressure of a gas above a solution is proportional to its mole fraction in the solution, \( p=K_H\,x \). A larger K_H means a lower solubility at a given pressure. It explains why soft drinks are bottled under pressure and why deep-sea divers are given air diluted with helium.

Vapour Pressure and Raoult's Law

The vapour pressure of a liquid is the pressure of its vapour in equilibrium with the liquid. It depends only on the nature of the liquid and the temperature, not on the amount of liquid or the size of the container. A liquid boils when its vapour pressure equals the external pressure.
For a solution of volatile liquids A and B, Raoult's law gives partial pressures \( p_A=x_AP_A^\circ,\ p_B=x_BP_B^\circ \) and the total pressure \( P=P_B^\circ+x_A(P_A^\circ-P_B^\circ) \). The vapour is richer in the more volatile liquid, with mole fraction \( y_A=\frac{p_A}{P} \).
For a non-volatile solute, only the solvent contributes: \( P=x_1P_1^\circ \), and the relative lowering of vapour pressure equals the mole fraction of the solute: \( \frac{P^\circ-P}{P^\circ}=x_2=\frac{n_2}{n_1+n_2} \).

Ideal and Non-Ideal Solutions

An ideal solution obeys Raoult's law at all concentrations and temperatures, with ΔH(mix) = 0 and ΔV(mix) = 0. It forms when A–B attractions are about equal to A–A and B–B attractions (benzene + toluene, hexane + heptane).
Positive deviation: A–B attractions are weaker, so the vapour pressure is higher than the Raoult value, ΔH(mix) > 0 and ΔV(mix) > 0 (ethanol + water, ethanol + acetone). Negative deviation: A–B attractions are stronger, so the vapour pressure is lower, ΔH(mix) < 0 and ΔV(mix) < 0 (chloroform + acetone, HNO₃ + water).
Azeotropes are constant-boiling mixtures that cannot be separated by distillation. Positive deviation gives a minimum boiling azeotrope (ethanol + water, about 95.5% ethanol); negative deviation gives a maximum boiling azeotrope (HNO₃ + water, about 68% HNO₃).

Colligative Properties

Properties that depend only on the number of solute particles and not on their nature are colligative: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.
1. Elevation of boiling point: \( \Delta T_b=K_bm \), where \( K_b=\frac{RT_b^2M_1}{1000\,\Delta H_{vap}} \) for the solvent (M₁ is its molar mass in g mol⁻¹).
2. Depression of freezing point: \( \Delta T_f=K_fm \). For a solute of mass w in solvent of mass W (grams): \( M_2=\frac{1000\,K\,w}{\Delta T\,W} \).
3. Osmotic pressure: \( \pi=CRT \) (R = 0.0821 L atm K⁻¹ mol⁻¹). Solutions with equal π at the same temperature are isotonic; a solution with higher π is hypertonic, with lower π hypotonic. Reverse osmosis, applying a pressure greater than π to the solution, is used to desalinate sea water.

Abnormal Molar Masses and van't Hoff Factor

When a solute associates or dissociates, the observed colligative effect differs from the calculated one. The van't Hoff factor is \( i=\frac{\text{observed colligative property}}{\text{calculated value}}=\frac{\text{normal molar mass}}{\text{observed molar mass}}=\frac{\text{total moles after}}{\text{moles before}} \).
The colligative equations become \( \Delta T_b=iK_bm,\ \Delta T_f=iK_fm,\ \pi=iCRT,\ \frac{\Delta P}{P^\circ}=ix_2 \) (approximately).
Dissociation into n ions with degree of dissociation α: \( i=1+(n-1)\alpha \) (i > 1). Association of n molecules into one with degree of association α: \( i=1-\alpha+\frac{\alpha}{n} \) (i < 1). Benzoic acid in benzene dimerises, so its observed molar mass is about double the formula mass.

Worked Example 1. Find the freezing point of a solution of 0.1 mol of NaCl in 1 kg of water, taking NaCl as 90% ionised (Kf = 1.86 K kg mol⁻¹).
Answer:
i = 1 + α = 1.9; m = 0.1 mol kg⁻¹.
ΔTf = 1.9 × 1.86 × 0.1 = 0.353 K, so the solution freezes at −0.353 °C.