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If the elevation in boiling point of a solution of $10 \mathrm{~g}$ of solute $(\mathrm{mol} . \mathrm{wt}=100)$ in $100 \mathrm{~g}$ of water is $\Delta T_{b}$, the ebullioscopic constant of water is
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The correct answer is:
$\Delta T_{b}$
We know that, $\Delta \mathrm{T}_{\mathrm{b}}=\frac{1000 \times \mathrm{K}_{\mathrm{b}} \times \mathrm{w}}{\mathrm{W} \times \mathrm{M}}$
$\begin{array}{l}
\mathrm{M}=\frac{1000 \times \mathrm{K}_{\mathrm{b}} \times \mathrm{w}}{\mathrm{W} \times \Delta \mathrm{T}_{\mathrm{b}}} \\
\Delta \mathrm{T}_{\mathrm{b}}=\frac{1000 \times \mathrm{K}_{\mathrm{b}} \times 10}{100 \times 100} \\
\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}}
\end{array}$
$\begin{array}{l}
\mathrm{M}=\frac{1000 \times \mathrm{K}_{\mathrm{b}} \times \mathrm{w}}{\mathrm{W} \times \Delta \mathrm{T}_{\mathrm{b}}} \\
\Delta \mathrm{T}_{\mathrm{b}}=\frac{1000 \times \mathrm{K}_{\mathrm{b}} \times 10}{100 \times 100} \\
\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}}
\end{array}$
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