The elevation in boiling point for a 0.1 m s | vedclass

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(D) The elevation in boiling point is given by the formula $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.For the first solutio

Vedclass · Accepted Answer

$0.51$

Vedclass · Answer

(D) The elevation in boiling point is given by the formula $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.For the first solution: $0.51 = K_b 	imes 0.1$,so $K_b = 5.1 \ K \cdot kg \cdot mol^{-1}$.When we mix two solutions of the same molality $(0.1 \ m)$,the resulting solution will also have a molality of $0.1 \ m$ because molality is an intensive property.Since the molality $(m)$ of the resulting solution remains $0.1 \ m$,the elevation in boiling point $\Delta T_b$ will remain $K_b 	imes 0.1 = 5.1 	imes 0.1 = 0.51 \ K$.

The elevation in boiling point for a $0.1 \ m$ solution of a solute in $500 \ g$ of benzene is $0.51 \ K$. If $1000 \ g$ of a $0.1 \ m$ solution of the same solute in benzene is added to the above solution,the $\Delta T_b$ for the resulting solution will be ............... $K$.

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