Question

The mole fraction of a solute in a binary solution is $0.1$. At $298K$, molarity of this solution is same as as its molality. Density of this solution at $298K$ is $2.0gc{m}^{-3}$. 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)

$X_2=0.1$
$\Rightarrow \frac{n_2}{n_2+n_1}=0.1\dots \dots (1)$
$M=\frac{W_2\times 1000}{M_2\times W}\times d\dots \dots (2)$
$m=\frac{W_2\times 1000}{M_2\times W_1(g)}\dots \dots (3)$
$d=2g/cc\dots (4)$
Given $M=m($ numerically $)$
$\Rightarrow \frac{W_2\times 1000}{M_2\times \left(W_2+W_1\right)}\times d=\frac{W_2\times 1000}{M_2\times W_1(g)}$
$\Rightarrow W_2+W_1=W_1\times d=2W_1$
$\Rightarrow W_2=W_1\dots \dots (5)$
Now, $\frac{n_2}{n_2+n_1}=0.1$
$\Rightarrow \frac{\frac{w_2}{M_2}}{\frac{W_2}{M_2}+\frac{W_1}{M_1}}=0.1$
$\Rightarrow \frac{\frac{W_2}{M_2}}{w_2\left(\frac{1}{M_2}+\frac{1}{M_1}\right)}=0.1$
$\Rightarrow \frac{\frac{1}{M_2}}{\frac{M_1+M_2}{M_1{M}_2}}=0.1$
$\Rightarrow \frac{M_1}{M_1+M_2}=\frac{1}{10}$
$\Rightarrow \frac{M_1}{M_1+M_2-M_1}=\frac{1}{10-1}$
$\Rightarrow \frac{M_1}{M_2}=\frac{1}{9}$
$\Rightarrow \frac{M_2}{M_1}=9$