Dry air was passed successively through a solution of 5g of a solute in 180g of water and then through pure water and then through pure water. The loss in weight of solution was `2.5g` and that of pure solvent `0.04g`. The molecular weight of the solute is:

Options

A. 50
B. 180
C. 100
D. 25

Answer

C

Explanation

Lowering in weight of solution `prop` solution pressure
Lowering in weight of solvent `prop P^(@)-P_(s)`
(`because p^(@)` = vapour pressure of pure solvent)
`(P^(@)-P_(s))/(P_(s))=("Lowering in weight of solvent")/("Lowering in weight of solution")`
`(P^(@)-P_(s))/(P_(s))=(w xx M)/(m xx W)`
`(0.05)/(2.5)=(10xx18)/(90xx m)rArr (2xx2.5)/(0.05)=(2xx250)/(5)=100`