If $0.01 \mathrm{~m}$ aqueous solution of an electrolyte freezes at $-0.056{ }^{\circ} \mathrm{C}$.…
If $0.01 \mathrm{~m}$ aqueous solution of an electrolyte freezes at $-0.056{ }^{\circ} \mathrm{C}$. Calculate van't Hoff factor for an electrolyte. (Cryoscopic constant of water $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
1.30
2.33
3.00
4.11
Solution
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \mathrm{m}$
$0.056=\mathrm{i} \times 1.86 \times 0.01$
$\therefore \quad \mathrm{i}=\frac{0.056}{1.86 \times 0.01}=3.01$
[Note: In the question, the freezing point of aqueous solution is changed from $-0.056 \mathrm{~K}$ to $-0.056^{\circ} \mathrm{C}$ to apply appropriate textual concepts.]