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_(A)^(@)P_(S))/P_(S)=(W_(B)xxW_(A))/(W_(A)xxM_(B))`
`((185-183))/((183))=((1.2 g)xx(58 g mol^(-1)))/((100g)xxM_(B))`
`M_(B)=((1.2g)xx(58g mol^(-1))xx(183))/(2xx(100g))=63.68g mol^(-1)`
=64 g `mol^(-1)`