Question

An aqueous solution of an electrolyte AB has b.pt.of 101.08$^\circ$C. The solute is 100% ionised at b.pt.of water. The f.pt. of the same solution is $-1 -80 ^\circ C. AB (K_bI/K_f = 0.3)$

• A. is 100% ionised at f.pt. of solution
• B. is 50% ionised at f.pt. of solution
• C. behaves as a non-electrolyte at f.pt. of solution
• D. forms a dimer at f.pt. of solution

Answer 1

Answer: (C)

behaves as a non-electrolyte at f.pt. of solution

Answer 2

$\Delta T_b = 1.08 \, K , \Delta T_f = 1.80 \, K$ $\frac{\Delta T_b}{\Delta T_f} = \frac{i. K_bm}{K_f . m}$ $\frac{\Delta T_b}{\Delta T_f} = \frac{i}{i'} \left(\frac{K_b}{K_f} \right)$ $i'$ = van't-Hoff factor at f.p.t of solution Here $i$ (at b.pt.) = 2 $\frac{1.08}{1.08} = \frac{2}{i'} (0.3)$ $i ' = \frac{2 \times 0.3 \times 1.80}{1.08 }$ = 1 Th is, solute is unionised at $f.pt$. of solution $i.e.$, behaves as a non-electrolyte