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(C) The elevation in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{\circ} = 100.18^{\circ} C - 100^{\circ} C = 0.18 \ K$. Since $\Delta T_{

Vedclass · Accepted Answer

$-0.64$

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(C) The elevation in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{\circ} = 100.18^{\circ} C - 100^{\circ} C = 0.18 \ K$. Since $\Delta T_{b} = K_{b} 	imes m$,the molality $m = \frac{\Delta T_{b}}{K_{b}} = \frac{0.18}{0.52} \ mol \ kg^{-1}$. The depression in freezing point is given by $\Delta T_{f} = K_{f} 	imes m$. Substituting the value of $m$: $\Delta T_{f} = 1.86 	imes \frac{0.18}{0.52} \approx 0.644 \ K$. The freezing point of the solution is $T_{f} = T_{f}^{\circ} - \Delta T_{f} = 0^{\circ} C - 0.644^{\circ} C = -0.644^{\circ} C$. Rounding to two decimal places,the freezing point is $-0.64^{\circ} C$.

$A$ solution of urea in water has a boiling point of $100.18^{\circ} C$. What is the freezing point of the same solution,if $K_{f}$ and $K_{b}$ of water are $1.86$ and $0.52 \ K \ kg \ mol^{-1}$,respectively (in $^{\circ} C$)? (Boiling point of water $= 100^{\circ} C$ )

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