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The mole fraction of a solute in a binary solution is $0.1 .$ At $298 \mathrm{~K}$, molarity of this solution is same as as its molality. Density of this solution at $298 \mathrm{~K}$ is $2.0 \mathrm{~g} \mathrm{~cm}^{-3}$. The ratio of molecular weights of the solute and the solvent $\left(\mathrm{M}_{\text {solute }} / \mathrm{M}_{\text {solvent }}\right)$ is
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9
$\mathrm{X}_{2}=0.1 \Rightarrow \frac{\mathrm{n}_{2}}{\mathrm{n}_{2}+\mathrm{n}_{1}}=0.1$ $\ldots(1)$
$\mathrm{M}=\frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}} \times \mathrm{d}$ $\ldots(2)$
$\mathrm{m}=\frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}_{1}(\mathrm{~g})}$ $\ldots(3)$
$\mathrm{d}=2 \mathrm{~g} / \mathrm{cc}$ $\ldots(4)$
Given M = m (numerically)
$\begin{array}{l}
\Rightarrow \frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times\left(\mathrm{W}_{2}+\mathrm{W}_{1}\right)} \times \mathrm{d}=\frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}_{1}(\mathrm{~g})} \\
\Rightarrow \mathrm{W}_{2}+\mathrm{W}_{1}=\mathrm{W}_{1} \times \mathrm{d}=2 \mathrm{~W}_{1} \\
\Rightarrow \mathrm{W}_{2}=\mathrm{W}_{1}
\end{array}$
Now, $\frac{n_{2}}{n_{2}+n_{1}}=0.1 \Rightarrow \frac{\frac{w_{2}}{M_{2}}}{\frac{W_{2}}{M_{2}}+\frac{W_{1}}{M_{1}}}=0.1 \Rightarrow \frac{\frac{W_{2}}{M_{2}}}{W_{2}\left(\frac{1}{M_{2}}+\frac{1}{M_{1}}\right)}=0.1$
$\Rightarrow \frac{\frac{1}{\mathrm{M}_{2}}}{\frac{\mathrm{M}_{1}+\mathrm{M}_{2}}{\mathrm{M}_{1} \mathrm{M}_{2}}}=0.1 \Rightarrow \frac{\mathrm{M}_{1}}{\mathrm{M}_{1}+\mathrm{M}_{2}}=\frac{1}{10} \Rightarrow \frac{\mathrm{M}_{1}}{\mathrm{M}_{1}+\mathrm{M}_{2}-\mathrm{M}_{1}}=\frac{1}{10-1} \Rightarrow \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\frac{1}{9} \Rightarrow \frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}=9$
$\mathrm{M}=\frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}} \times \mathrm{d}$ $\ldots(2)$
$\mathrm{m}=\frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}_{1}(\mathrm{~g})}$ $\ldots(3)$
$\mathrm{d}=2 \mathrm{~g} / \mathrm{cc}$ $\ldots(4)$
Given M = m (numerically)
$\begin{array}{l}
\Rightarrow \frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times\left(\mathrm{W}_{2}+\mathrm{W}_{1}\right)} \times \mathrm{d}=\frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}_{1}(\mathrm{~g})} \\
\Rightarrow \mathrm{W}_{2}+\mathrm{W}_{1}=\mathrm{W}_{1} \times \mathrm{d}=2 \mathrm{~W}_{1} \\
\Rightarrow \mathrm{W}_{2}=\mathrm{W}_{1}
\end{array}$
Now, $\frac{n_{2}}{n_{2}+n_{1}}=0.1 \Rightarrow \frac{\frac{w_{2}}{M_{2}}}{\frac{W_{2}}{M_{2}}+\frac{W_{1}}{M_{1}}}=0.1 \Rightarrow \frac{\frac{W_{2}}{M_{2}}}{W_{2}\left(\frac{1}{M_{2}}+\frac{1}{M_{1}}\right)}=0.1$
$\Rightarrow \frac{\frac{1}{\mathrm{M}_{2}}}{\frac{\mathrm{M}_{1}+\mathrm{M}_{2}}{\mathrm{M}_{1} \mathrm{M}_{2}}}=0.1 \Rightarrow \frac{\mathrm{M}_{1}}{\mathrm{M}_{1}+\mathrm{M}_{2}}=\frac{1}{10} \Rightarrow \frac{\mathrm{M}_{1}}{\mathrm{M}_{1}+\mathrm{M}_{2}-\mathrm{M}_{1}}=\frac{1}{10-1} \Rightarrow \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\frac{1}{9} \Rightarrow \frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}=9$
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