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

`(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))=(185-183)/(185)=(1.2//M_(2))/(100//58)`
`("Molar mass of "CH_(3)COCH_(3)="58 g mol"^(-1))`
`=(1.2xx58)/(100xxM_(2)) or M_(2)=(1.2xx58)/(100)xx(185)/(2)`
`="64.38 g mol"^(-1)`