Question

The mole fraction of a solute in a binary solution is $0.1$. At $298 \,K$, molarity of this solution is same as as its molality. Density of this solution at $298\, K$ is $2.0\, g \,cm ^{-3}$. The ratio of molecular weights of the solute and the solvent $\left(M_{\text {solute }} / M_{\text {solvent }}\right)$ is

• A. $9$
• B. $\frac{1}{9}$
• C. $4.5$
• D. $\frac{1}{4.5}$

Answer 1

Answer: (A)

$X_{2}=0.1 $
$\Rightarrow \frac{n_{2}}{n_{2}+n_{1}}=0.1 \ldots \ldots (1)$
$M=\frac{W_{2} \times 1000}{M_{2} \times W} \times d \ldots \ldots (2)$
$m=\frac{W_{2} \times 1000}{M_{2} \times W_{1}(g)} \ldots \ldots (3)$
$d=2 g / c c \ldots(4)$
Given $M = m ($ numerically $)$
$\Rightarrow \frac{W_{2} \times 1000}{M_{2} \times\left(W_{2}+W_{1}\right)} \times d=\frac{W_{2} \times 1000}{M_{2} \times W_{1}(g)}$
$\Rightarrow W_{2}+W_{1}=W_{1} \times d=2 W_{1}$
$\Rightarrow W_{2}=W_{1} \ldots \ldots (5)$
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}{M_{2}}}{\frac{M_{1}+M_{2}}{M_{1} M_{2}}}=0.1 $
$\Rightarrow \frac{M_{1}}{M_{1}+M_{2}}=\frac{1}{10}$
$ \Rightarrow \frac{M_{1}}{M_{1}+M_{2}-M_{1}}=\frac{1}{10-1}$
$\Rightarrow \frac{M_{1}}{M_{2}}=\frac{1}{9}$
$ \Rightarrow \frac{M_{2}}{M_{1}}=9$