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(C) The formula for boiling point elevation is $\Delta T_{b} = K_{b} 	imes m$,where $\Delta T_{b}$ is the elevation in boiling point,$K_{b}$ is the e

Vedclass · Accepted Answer

$0.385 \ mol \ kg^{-1}$

Vedclass · Answer

(C) The formula for boiling point elevation is $\Delta T_{b} = K_{b} 	imes m$,where $\Delta T_{b}$ is the elevation in boiling point,$K_{b}$ is the ebullioscopic constant,and $m$ is the molality of the solution. Given: $\Delta T_{b} = 0.2 \ K$ and $K_{b} = 0.52 \ K \ kg \ mol^{-1}$. Rearranging the formula to solve for molality $(m)$: $m = \frac{\Delta T_{b}}{K_{b}}$. Substituting the values: $m = \frac{0.2}{0.52} \approx 0.3846 \ mol \ kg^{-1}$. Rounding to three decimal places,we get $m \approx 0.385 \ mol \ kg^{-1}$. Therefore,the correct option is $C$.

Calculate the molality of a solution containing a non-volatile solute if the boiling point elevation of the solution is $0.2 \ K$ $[K_{b} = 0.52 \ K \ kg \ mol^{-1}]$.

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Give the formula for the boiling point elevation constant $(K_b)$.

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Elevation in boiling point was $0.52 \, ^\circ C$ when $6 \, g$ of a compound $X$ was dissolved in $100 \, g$ of water. The molecular weight of $X$ is: ($K_b$ for water is $0.52 \, K \, kg \, mol^{-1}$)

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