7g of a solute are dissolved in 200 g of a solvent (mol. Wt. 78) to prepare a solution. At a certain temperature dry air is sent into a solution and solvent continuously. The loss is weights of containers containing solutions and solvent are 0.975 g and 0.025 g. Calculate the molar mass of solute.

Explanation

Lowering of vapour pressure `(P^0 - P)` is proportional to `y , y = 0.025 g`
Vapour pressure (P) is proportional to `x , x = 0.975 g`
Vapour pressure of the pure solvent = `p^0 = x + 4 = 0.975 + 0.025 = 1g`
`y/(x + y) = (p^0 - p)/(p^0) = (wM)/(Wm) = (0.025)/(1.0)`
Molar mass of solute, `m = (7 xx 78)/(0.025 xx 200) = 100 g mol^(-1)`.