If 0.15 m aqueous solution of KCl freezes at | vedclass

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(C) The depression in freezing point is given by $\Delta T_{f} = T_{f}^0 - T_{f}$.Given $T_{f}^0 = 0^{\circ} C$ and $T_{f} = -0.51^{\circ} C$,so $\Del

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$1.82$

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(C) The depression in freezing point is given by $\Delta T_{f} = T_{f}^0 - T_{f}$.Given $T_{f}^0 = 0^{\circ} C$ and $T_{f} = -0.51^{\circ} C$,so $\Delta T_{f} = 0 - (-0.51) = 0.51 \ K$.The formula for depression in freezing point is $\Delta T_{f} = i 	imes K_{f} 	imes m$.Substituting the values: $0.51 = i 	imes 1.86 	imes 0.15$.Solving for $i$: $i = \frac{0.51}{1.86 	imes 0.15} = \frac{0.51}{0.279} \approx 1.828$.Rounding to two decimal places,the van't Hoff factor $i$ is $1.82$.

If $0.15 \ m$ aqueous solution of $KCl$ freezes at $-0.51^{\circ} C$,calculate the van't Hoff factor of $KCl$ (cryoscopic constant of water is $1.86 \ K \ kg \ mol^{-1}$).

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