Dry air was passed successively through a solution of 5 g of a solute in 80 g of water and then through pure water. The loss in mass of solution was 2.5 g and that of pure solvent was 0.04 g. What is

Question

Dry air was passed successively through a solution of 5 g of a solute in 80 g of water and then through pure water. The loss in mass of solution was 2.5 g and that of pure solvent was 0.04 g. What is the molecular mass of the solute?

Prashant Kotian · Accepted Answer

Correct answer: A. According to the Ostwald-Walker (dynamic method): Loss of mass of solution bulbs (w_sol) ∝ P, and Loss of mass of solvent bulb (w_solv) ∝ P^o - P. Thus: (P^o - P) / P = w_solv / w_sol = 0.04 / 2.5 = 0.016. According to Raoult's law: (P^o - P) / P = n_solute / n_solvent = (5 / M) / (80 / 18). So, 0.016 = (5 / M) * (18 / 80) = 90 / (80 M) = 1.125 / M => M = 1.125 / 0.016 ≈ 70.31.

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