The vapor pressure of acetone at `20^(@)C` is 185 torr. When `1.2 g` of a non-volatile solute was dissolved in `100 g` of acetone at `20^(@)C`, it vapour pressure was 183 torr. The molar mass `(g mol^(-1))` of solute is:

Options

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

Answer

D

Explanation

According to Raoult's law
`(P_(0)-P_(s))/(P_(s)) =(n)/(N) rArr (2)/(183) = (((1.2)/(M)))/(((100)/(58)))`
`rArr M = 64`