`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(wxx1000)/(mxxW)`
where, `Delta T` = Depression in freezing point
`K_(f)=` Molal depression constant of benzene
w = Mass of solute
m = Molecular mass of solute
W = Mass of solvent
Case I : `(278.4-277.12)=K_(f)xx(1.4xx1000)/(58.100)`
`1.28=K_(f)xx(14)/(58)`....(i)
Case II : `(278.4-277.76)=K_(f)xx(2.8xx100)/(m_((A))xx100)`
`0.64=K_(f)(28)/(m_((A)))`....(ii)
Dividing eq. (i) by eq. (ii), we get
`m_((A))=232`.