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).
Answer & explanation
Correct answer: option 3
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, \(m_{(A)} = 232\)