Dissociation of salt A2B3(aq) in water follows first-order kinetics: A2B3(aq) → 2A^3+(aq) + 3B^2-(aq). Starting from t = 0, a definite volume of solution is taken at different instant and its osmotic

Question

Dissociation of salt A2B3(aq) in water follows first-order kinetics: A2B3(aq) → 2A^3+(aq) + 3B^2-(aq). Starting from t = 0, a definite volume of solution is taken at different instant and its osmotic rise is measured. If the osmotic rise was 2 mm at t = 0 and 6 mm at t = 10 min, then the value of x/y is, where x = osmotic pressure of solution (in Pascal) at t = 20 min and y = t1/4 of the dissociation reaction (in min) [Given: Density of final solution in each case = 1.0 g/ml, g = 10 m/s^2]

Prashant Kotian · Accepted Answer

Correct answer: 4. Initial rise h0 = 2 mm (proportional to C0). Total ions = 5. At time t, h_t = C0(1 + 4α). At 10 min, h = 6 mm, so 2(1 + 4α) = 6 => α = 0.5. Half-life is 10 min. At 20 min (2 half-lives), α = 0.75. h_20 = 2(1 + 4(0.75)) = 8 mm. Osmotic pressure x = ρgh = (1000 kg/m^3)(10 m/s^2)(8 × 10^-3 m) = 80 Pascal. y is the time to reduce to 1/4th concentration (2 half-lives), so y = 20 min. x/y = 80 / 20 = 4. Therefore, correct answer is 4.

💡 Solution & Explanation

Practice 22,000+ questions like this