`1.4 g` of acetone dissolved in `100 g` of benzene gave a solution which freezes at `277.12 K`. Pure benzene freezes at `278.4 K`.`2.8` of solid `(A)` dissolved in `100 g` of benzene gave a solution which froze at `277.76 K`. Calculate the molecular mass of `(A)`.

Explanation

For acetone `+` Benzene mixture:
`Delta T = (K'_(f)xx1000 xxw)/(mxxW)`
`(278.40-277.12) = (1000 xx K_(f) xx 1.4)/(100 xx 58)`
or `1.28 = (1000 xx K_(f) xx 1.4)/(100 xx 58)` …(1)
For solute `(A) +` Benzene mixture, (Let `m` be the mol. weight of `A`)
`(278.40-277.76) = (1000 xx K_(f) xx 2.8)/(100 xx m)`
or `0.64 = (1000 xx K_(f) xx 2.8)/(100 xx m)` ...(2)
By eqs. (1) and (2),
`m = 232`