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$5 \mathrm{~g}$ of non-volatile water soluble compound $X$ is dissolved in $100 \mathrm{~g}$ of water. The elevation in boiling point is found to be 0.25 . The molecular mass of compound $X$ is
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The correct answer is:
20 g
The elevation in boiling point is,
$\begin{aligned}
\Delta T_b & =\frac{1000 k_b \times w_2}{M_2 \times w_1(\text { in }(\mathrm{g}))} \Rightarrow M_2=\frac{1000 \times k_b \times 5}{0.25 \times 100} \\
\Rightarrow M_2 & =20 \mathrm{~g}
\end{aligned}$
$\begin{aligned}
\Delta T_b & =\frac{1000 k_b \times w_2}{M_2 \times w_1(\text { in }(\mathrm{g}))} \Rightarrow M_2=\frac{1000 \times k_b \times 5}{0.25 \times 100} \\
\Rightarrow M_2 & =20 \mathrm{~g}
\end{aligned}$
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