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

`((6 xx1000)/(60xx100))_("urea") = C_("glucose")`
`therefore " " C_("glucose") = 1 M`
`(p^(@) -p_(s))/(p^(@)) ~~ x_(B) = (n_(B))/(n_(A) + n_(B)) ~~(n_(B))/(n_(A))`
`~~(w_(B) xx M_(A))/(w_(A) xx M_(B))`
`(185-183)/(185) = (1.2 xx 58 )/(100 xx M_(2))`
`therefore " "M_(2) = (1.25 xx 58 xx 185)/(2 xx 100) = 64.38 ~~ 64`