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

232

The correct answer is option 3. 232

We know that,

\(\Delta T = K_f × \frac{w × 1000}{m × W}\)

where, \(\Delta T\text{ = Depression in freezing point}\)

\(K_f \text{ = Molal depression constant of benzene}\)

\(w \text{ = Mass of solute}\)

\(m \text{ = Molecular mass of the solute}\)

\(W \text{ = Mass of solvent}\)

Case I:

\((278.4 − 277.12) = K_f × \frac{1.4 × 1000}{58 × 100}\)

or, \(1.28 = K_f × \frac{14}{58}\)------(i)

Case II:

\((278.4 − 277.76) = K_f × \frac{2.8 × 1000}{m_{(A)} × 100}\)

or, \(0.64 = K_f × \frac{28}{m_{(A)}}\)------(ii)

Dividing equation (i) by (ii), we get

\(\frac{1.28}{0.64} = \frac{14/58}{28/m_{(A)}}\)

or, \(2 = \frac{m_{(A)}}{2}\)

or, \(m_{(A)} = 232\)