`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

We know that
`DeltaT = (K_(f)) xx (W_(2) xx 1000)/(Mw_(2) xx W_(1))`
where`DeltaT` = Depression in freezing point
`K_(f)` = Molal depression constant of benzene`
`W_(2)` = Mass of solute
`Mw_(2)` = Molecular mass of solute
`W_(1)` = Mass of solvent
Case i: `(278.4 - 277.12) = K_(f)xx(1.4 xx 1000)/(58 xx 100)`
`1.28 = K_(f) xx (14)/(58)`
Case ii: `(278.4 - 277.76) = K_(f)xx(2.8 xx 1000)/(Mw_(A) xx 100)`
`0.64 = K_(f) xx (28)/(Mw_(A))`
Dividing Eq. (i) by (ii), we get
`Mw_(A) = 232`