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.