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

Options

A. 31.25
B. 3.125
C. 312.5
D. None

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 180)`
`:. Mw_(2) =31.25`