Elevation in boiling point for 1.5 , m soluti | vedclass

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Question

(D) For glucose,the van't Hoff factor $i = 1$.The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m_{1} = 4 \, K$,where $m_{1} = 1.

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) For glucose,the van't Hoff factor $i = 1$.The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m_{1} = 4 \, K$,where $m_{1} = 1.5 \, m$.So,$K_{b} \cdot 1.5 = 4 \implies K_{b} = \frac{4}{1.5}$.The depression in freezing point is given by $\Delta T_{f} = K_{f} \cdot m_{2} = 4 \, K$,where $m_{2} = 4.5 \, m$.So,$K_{f} \cdot 4.5 = 4 \implies K_{f} = \frac{4}{4.5}$.The ratio $\frac{K_{b}}{K_{f}} = \frac{4 / 1.5}{4 / 4.5} = \frac{4.5}{1.5} = 3$.

Elevation in boiling point for $1.5 \, m$ solution of glucose in water is $4 \, K$. The depression in freezing point for $4.5 \, m$ solution of glucose in water is $4 \, K$. The ratio of molal elevation constant to molal depression constant $(K_{b} / K_{f})$ is $....$

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