Question

An aqueous solution of a solute $AB$ has $b.pt.$ of $101.08^{\circ}C$ ($AB$ is $100\%$ ionised at boiling point of the solution) and freezes at $-1.80^{\circ}C$. Hence $AB$ ($K_{b}/K_{f} = 0.3$)

• A. is $100\%$ ionised at the f.pt. of the solution
• B. behave as non-electrolyte at f.pt. of solution
• C. forms dimer
• D. none of the above

Answer 1

Answer: (B)

Given, $\Delta T_{b} = 1.08^{\circ}C, i = 2$ at b.pL of solution,
$\Delta T_{f} = 180^{\circ}C$ and $\frac{K_{b}}{K_{f}} = 0.3$
So $\frac{\Delta T_{b}}{\Delta T_{f}} = \frac{i_{b} K_{b}m}{i_{f} K_{f}m}$
$\Rightarrow i_{f} = \frac{i_{b} K_{b}m}{K_{f} m} \times \frac{\Delta T_{f}}{\Delta T_{b}}$
and $i_{f} = 1$
i.e., $AB$ behave as non-electrolyte at f.pt. of solution.