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

Using relation,
`(p^@-p_s)/p_S = (w_2M_1)/(w_1M_2)`
where `w_1, M_1` = mass in g and mol. Mass of solvent `w_2, M_2`= mass in g and mol. Mass of solute
Let `M_2=x`
`p^@ = 185" torr"`
`p_s = 183" torr"`
`(185 -183)/183=(1.2 xx 58) /(100x)` (Mol. mass of acetone = 58)
`x = 64`
`:.`Molar mass of substance = 64