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. 32
B. 64
C. 128
D. 488

Answer

B

Explanation

`P_0 = 185 ` torr , `P_S = 183 ` torr
w= 1.2 g , W = 100 g
M of `CH_3 COCH_3 = 58 , m = ` ?
`(P_0-P_s)/(P_0)=n/(n+N)`
`(185-183)/(185)=(1.2 //m)/(100//58)" "[nltltN]`
m = 64.55