The amount of solute (molar mass 60 g `mol^(-1)`) that must be added to 180 g of water so that the vapour pressure of water is lowered by `10%` is

Options

A. 30 g
B. 60 g
C. 120 g
D. 12 g

Answer

B

Explanation

Relative lowering of vapour pressure is given by the formula :
`(p^(@)-p_(s))/(p^(@))=(omega_(A))/(m_(A))xx(m_(B))/(omega_(B))`
Given, relative lowering of vapour pressure
`=(p^(@)-p_(s))/(p^(@))=(10)/(100)`
`m_(A)=60, m_(B)=18, omega_(B)=180, omega_(A)=x`
`therefore (10)/(100)=(x//60)/(180//18)rArr(1)/(10)=(x//60)/(10)rArr x = 60`
Thus, 60 g of the solute must be added to 180 g of water so that the vapour pressure of water is lowered by 10%.