Dry air was passed successively through a solution of 5 g of a solute in 180 g of water and then through pure water. The loss in the weight of solution was 2.50 g and that of pure solvent 0.04 g. The molecular weight of the solute is.

Options

A. `31.25`
B. `3.125`
C. `312.5`
D. `0.3125`

Answer

A

Explanation

`P^(@)- P_(S) prop` loss in weight of water chamber and `P_(S) prop` loss in weight of solution chamber.
`(P^(@)-P_(S))/(P_(S))= (n_(2))/(n_(1)) (W_(2)xx Mw_(1))/(Mw_(2) xx W_(1)) or (0.04)/(2.50)= (5 xx 18)/(Mw_(2)xx W_(1))`
`therefore Mw_(2)=31.25`