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

Options

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

Answer

D

Explanation

Using relation,
`(p^(@) -p_(s))/(p _(s))=(w_(2) M_(1))/(w_(1)M_(2))`
`w_(2), M_(2)=` mas in g and mol mass of solvcnt `w_(2), M _(2)=` mass in g and mol. Mass of solutc
Let `M_(2) =x`
`p^(@)=185` torr
`p _(s) =183 ` torr
`(185-183)/(183)=(1.2xx58)/(100x)` (mol. mass of acctone = 58)`
`x= 64`
`therefore ` Molar mass of substance = 64