If $0.15 \mathrm{~m}$ aqueous solution of $\mathrm{KCI}$ freezes at $-0.51^{\circ} \mathrm{C}$, calculate…

If $0.15 \mathrm{~m}$ aqueous solution of $\mathrm{KCI}$ freezes at $-0.51^{\circ} \mathrm{C}$, calculate van't Hoff factor of $\mathrm{KCI}$ (cryoscopic constant of water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
  1. $1.45$
  2. $1.26$
  3. $1.82$
  4. $3.00$

Solution

$\begin{array}{ll} & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{f}}^0-\mathrm{T}_{\mathrm{f}}=0-\left(-0.51^{\circ} \mathrm{C}\right)=0.51{ }^{\circ} \mathrm{C}=0.51 \mathrm{~K} \\ \therefore & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{f}} \mathrm{m} \\ \therefore & \mathrm{i}=\frac{\Delta \mathrm{T}}{\mathrm{K}_{\mathrm{f}} \mathrm{m}}=\frac{0.51 \mathrm{~K}}{1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 0.15 \mathrm{~mol} \mathrm{~kg}^{-1}}=1.82\end{array}$

Asked in: MHT CET 2023 (12 May Shift 2)

Practice more Solutions questions on Aicharya