Decomposition of a non-volatile solute 'A' into another non-volatile solute 'B' and 'C', in aqueous solution follows first-order kinetics as: A → 2B + C. When one mole of 'A' is dissolved in 180 g wat

Question

Decomposition of a non-volatile solute 'A' into another non-volatile solute 'B' and 'C', in aqueous solution follows first-order kinetics as: A → 2B + C. When one mole of 'A' is dissolved in 180 g water and left for decomposition, the vapour pressure of solution was found to be 20 mm Hg after 12 h. What is the vapour pressure of solution after 24 h? Assume constant temperature of 25°C, throughout. The vapour pressure of water at 25°C is 24 mm Hg.

Prashant Kotian · Accepted Answer

Correct answer: B. Moles of H2O = 180/18 = 10. Initially 1 mole A. After 12h, P = 24 × (10 / (10 + 1 + 2x)) = 20 => 240 = 20(11+2x) => x = 0.5. So 12h is the half-life. After 24h (2 half-lives), x = 0.75. Total solute moles = 1 + 2(0.75) = 2.5. P = 24 × (10 / 12.5) = 19.2 mm Hg. Therefore, correct answer is B.

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