JEE Chemistry Solutions Notes Set 01

Types of Solutions and Concentration Terms

A solution is a homogeneous mixture; the component present in larger amount is the solvent. Common ways to express concentration:

Useful relations: \( M = \dfrac{10\times d\times \%(w/w)}{M_{solute}} \); the sum of mole fractions is 1; pure water is 55.5 M. For H₂O₂, volume strength = 11.2 × M.
Examples of n-factor: HCl 1, H₂SO₄ 2, H₃PO₄ 3, H₃PO₃ 2, Na₂CO₃ 2.

Worked example 1. Find the molarity of 98% H₂SO₄ by mass (density 1.8 g mL⁻¹).
Answer:
1000 mL has mass 1800 g, containing 0.98 × 1800 = 1764 g H₂SO₄.
Moles = 1764/98 = 18, so molarity = 18 M.

Solubility of Gases: Henry's Law

The partial pressure of a gas in the vapour phase is proportional to its mole fraction in solution: p = K_H x. A larger K_H means a lower solubility, and K_H rises with temperature, so gases are less soluble in warm water. Gases that react with the solvent (NH₃, HCl in water) do not obey the law.

Vapour Pressure and Raoult's Law

For a solution of volatile liquids A and B, p_A = x_AP_A⁰ and p_B = x_BP_B⁰; the total pressure is P = P_B⁰ + x_A(P_A⁰ − P_B⁰). The vapour is richer in the more volatile component, with y_A = p_A/P.
For a non-volatile solute, the relative lowering of vapour pressure is \( \dfrac{P^0-P_s}{P^0} = x_{solute} = \dfrac{n}{n+N} \) and \( \dfrac{P^0-P_s}{P_s} = \dfrac{n}{N} \).
A liquid boils when its vapour pressure equals the external pressure.

Ideal and Non-ideal Solutions

For an ideal solution, ΔH_mix = 0, ΔV_mix = 0, ΔS_mix > 0 and ΔG_mix < 0.

Colligative Properties

Colligative properties depend on the number of solute particles, not their nature: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.
\( \Delta T_b = K_b m,\qquad \Delta T_f = K_f m,\qquad \pi = CRT = \dfrac{n}{V}RT \), and the molar mass of a solute is \( M = \dfrac{K\times w\times 1000}{\Delta T\times W} \) (w = mass of solute, W = mass of solvent in g).
For water K_b = 0.52 and K_f = 1.86 K kg mol⁻¹; for benzene K_b = 2.53 and K_f = 5.12. K_b and K_f depend only on the solvent.
Osmosis is the net flow of solvent through a semipermeable membrane from the dilute to the concentrated solution. Solutions with equal osmotic pressure are isotonic. Osmotic pressure is the best method for molar masses of polymers and proteins because it is measurable at room temperature and low concentration.

Worked example 2. 68.5 g of sucrose (M = 342) is dissolved in 1000 g of water. Find the freezing point (K_f = 1.86).
Answer:
m = 68.5/342 = 0.2 mol kg⁻¹.
ΔT_f = 1.86 × 0.2 = 0.372, so the freezing point is −0.372°C.

Abnormal Colligative Properties and van't Hoff Factor

Dissociation or association of the solute changes the number of particles. The van't Hoff factor \( i = \dfrac{\text{observed colligative property}}{\text{calculated colligative property}} = \dfrac{M_{normal}}{M_{observed}} \).
For dissociation into n ions with degree α, i = 1 + (n − 1)α. For association of n molecules into one, i = 1 − α(1 − 1/n). Examples (complete dissociation): NaCl 2, CaCl₂ 3, K₂SO₄ 3, Al₂(SO₄)₃ 5, K₄[Fe(CN)₆] 5.
All colligative formulas are multiplied by i: ΔT_b = iK_bm, ΔT_f = iK_fm, π = iCRT. Acetic acid and benzoic acid dimerise in benzene (i about 0.5), but ionise slightly in water (i slightly above 1).

Worked example 3. The osmotic pressure of 0.1 M Na₂SO₄ is measured. If i = 2.5, find the degree of dissociation.
Answer:
i = 1 + 2α = 2.5, so α = 0.75, i.e. 75% dissociated.