The vapour pressure of acetone at `20^(@)C` is 185 torr. When `1.2 g` of non-volatile substance was dissolved in `100g` of acetone at `20^(@)C` its vapour pressure was 183 torr. The moalr mass `(g mol^(-1))` of the substance is:

Options

A. 32
B. 64
C. 128
D. 488

Answer

B

Explanation

Vapour pressure of pure acetone `P_(A)^(@)=185` torr
Vapour pressure of solution, `P_(S)=183` torr
Molar mass of solvent, `M_(A)=58` g//mol
as we know `(P_(A)^(@)-P_(S))/(P_(S))=(n_(B))/(n_(A))`
`rArr (185-183)/(183)=(W_(B))/(M_(B))xx(M_(A))/(W_(A))`
`rArr (2)/(183)=(1.2)/(M_(B))xx(58)/(100)`
`rArr M_(B)=(1.2)/(2)xx(58)/(100)xx183=64` g/mol