JEE Chemistry Solutions Notes Set 03

Types of Solutions and Solubility

A solution is a homogeneous mixture of two or more substances. The component present in the larger amount is the solvent and the others are solutes. Solutions can be gaseous, liquid or solid (alloys), depending on the physical state of the solvent.
Solubility is the maximum amount of solute that dissolves in a given amount of solvent at a given temperature (a saturated solution). Like dissolves like: polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents. For solids dissolving with absorption of heat, solubility rises with temperature; for gases in liquids, solubility falls as temperature rises and rises with pressure (Henry's law: the mass of gas dissolved is proportional to its pressure).

Ways of Expressing Concentration

Mass percentage = (mass of solute / mass of solution) × 100. Parts per million = (mass of solute / mass of solution) × 10⁶.
Mole fraction x = moles of the component / total moles; the mole fractions of all components add up to 1.
Molarity M = moles of solute per litre of solution. Normality N = gram equivalents of solute per litre of solution = M × n-factor. Molality m = moles of solute per kg of solvent. Formality is used for ionic solids that do not exist as molecules.
Molality and mole fraction do not change with temperature because they depend on masses; molarity and normality change because the volume changes.
Conversion: M = (10 × percentage by mass × density in g/mL) / molar mass. Dilution: M₁V₁ = M₂V₂. Mixing: M = (M₁V₁ + M₂V₂)/(V₁ + V₂). Neutralisation: N₁V₁ = N₂V₂.

Vapour Pressure and Raoult's Law

Vapour pressure is the pressure of the vapour in equilibrium with its liquid at a given temperature; it depends only on the nature of the liquid and the temperature.
For a solution of volatile liquids A and B, Raoult's law says p_A = x_A p°_A and p_B = x_B p°_B, and the total pressure is p = p_A + p_B. The mole fraction of A in the vapour is y_A = p_A / p.
For a non-volatile solute, the vapour pressure of the solution is p = x_solvent p°, and the relative lowering of vapour pressure is \( \frac{p^{\circ}-p}{p^{\circ}}=x_{solute}=\frac{n}{n+N} \), which for dilute solutions is about n/N = (w₂ M₁)/(M₂ w₁). This gives the molar mass of the solute.

Ideal and Non-Ideal Solutions, Azeotropes

An ideal solution obeys Raoult's law at all concentrations, with ΔH_mix = 0 and ΔV_mix = 0 because A–B forces equal A–A and B–B forces. Examples: benzene and toluene, n-hexane and n-heptane, ethyl bromide and ethyl iodide.
Positive deviation: A–B forces are weaker than A–A and B–B forces, so the vapour pressure is higher than expected, ΔH_mix > 0 and ΔV_mix > 0 (ethanol and water, acetone and carbon disulphide, benzene and methanol). Negative deviation: A–B forces are stronger, so the vapour pressure is lower, ΔH_mix < 0 and ΔV_mix < 0 (acetone and chloroform, nitric acid and water, HCl and water).
Azeotropes are mixtures that boil at constant temperature and composition, so they cannot be separated by fractional distillation. Positive deviation gives a minimum-boiling azeotrope (ethanol and water, 95.6% ethanol); negative deviation gives a maximum-boiling azeotrope (nitric acid and water, 68% HNO₃; HCl and water, about 20.2% HCl).

Colligative Properties

Properties that depend only on the number of solute particles and not on their nature are called colligative properties: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. Optical rotation, colour, density and viscosity are not colligative.
Elevation of boiling point: ΔT_b = K_b m, where K_b is the molal elevation constant (water 0.52 K kg/mol). Depression of freezing point: ΔT_f = K_f m (water 1.86, benzene 5.12). The molar mass follows from \( M_2=\frac{K\,w_2\times1000}{\Delta T\,w_1} \).
Osmosis is the flow of solvent through a semipermeable membrane from a dilute to a concentrated solution. Osmotic pressure π = CRT = nRT/V (R = 0.0821 L atm/(K mol)). Solutions of equal osmotic pressure are isotonic; a more concentrated one is hypertonic and a more dilute one hypotonic. Reverse osmosis, in which a pressure greater than π is applied, is used to desalinate sea water. Osmotic pressure is the best method for the molar mass of polymers and proteins because it gives a measurable value even for very dilute solutions at room temperature.

Abnormal Molar Masses and the Van't Hoff Factor

Solutes that dissociate (electrolytes) or associate (acetic acid or benzoic acid in benzene, which form dimers) give abnormal colligative properties. The Van't Hoff factor i = (observed colligative property) / (calculated value) = (normal molar mass) / (observed molar mass) = (moles of particles after) / (moles before).
For dissociation into n ions with degree of dissociation α: i = 1 + (n − 1)α. For association of n molecules into one: i = 1 − α + α/n (dimer: 1 − α/2). All colligative formulas are multiplied by i. Typical values at complete dissociation: NaCl 2, BaCl₂ 3, Al₂(SO₄)₃ 5, K₃[Fe(CN)₆] 4, urea and glucose 1.
Colligative effects of equimolar solutions rise with i, so for equal molarity the order of boiling point, osmotic pressure and vapour-pressure lowering is glucose < NaCl < Na₂SO₄ < Na₃PO₄ < Al₂(SO₄)₃, and the freezing points fall in the same order.

Solved Examples

Example 1. Calculate the molarity and molality of 98% H₂SO₄ (density 1.84 g/mL) by weight.
Answer:
Molarity = (10 × 98 × 1.84)/98 = 18.4 M.
In 100 g of acid: H₂SO₄ = 98 g (1 mol), water = 2 g, so molality = 1/0.002 = 500 m.

Example 2. The vapour pressure of pure benzene at 25 °C is 100 mm Hg. 1 g of a non-volatile solute in 39 g of benzene lowers it to 95 mm Hg. Find the molar mass of the solute (benzene = 78).
Answer:
Relative lowering = 5/100 = 0.05 = n/(n + 0.5) gives n = 0.0263 mol.
M = 1/0.0263 = 38 g/mol.

Example 3. Find the freezing point of a 0.1 m aqueous solution of K₂SO₄ assuming 100% dissociation (Kf = 1.86).
Answer:
i = 3, so ΔT_f = 3 × 1.86 × 0.1 = 0.558 K.
Freezing point = −0.558 °C.

Example 4. Calculate the osmotic pressure of a solution containing 6 g of urea in 500 mL at 27 °C.
Answer:
C = (6/60)/0.5 = 0.2 mol/L.
π = CRT = 0.2 × 0.0821 × 300 = 4.93 atm.