QJEEAdv16Chee86: JEE Advanced 2016 Chemistry - Class 12 JEEAdv16Chee86 Solution: The mole fraction of a solute in a solution is 0.1. At 298 K,…

It is given that the mole fraction of the solute $\left(x_{\text {solute }}\right)$ is 0.1 , therefore mole fraction of the solvent $\left(x_{\text {solvent }}\right)$ is 0.9 . So we have

$x_{\text {solute }}=\frac{n_{\text {solute }}}{\left(n_{\text {solute }}+n_{\text {solvent }}\right)}=0.1$ ......(1)

$x_{\text {solvent }}=\frac{n_{\text {solvent }}}{\left(n_{\text {solute }}+n_{\text {solvent }}\right)}=0.9$ ......(2)

where $n_{\text {solute }}$ and $n_{\text {solvent }}$ are the number of moles of solute and solvent, respectively. Dividing Eq. (1) by Eq. (2) gives

$\frac{x_{\text {solute }}}{x_{\text {solvent }}}=\frac{n_{\text {solute }}}{n_{\text {solvent }}}=\frac{0.1}{0.9}$ .......(3)

Given that the density of solution is $2 \mathrm{~g} \mathrm{~cm}^{-3}$ , we have

$\begin{aligned}& W_{\text {solution }} =\text { density } \times V_{\text {solution }}=2 \times V_{\text {solution }} \\\\& \therefore W_{\text {solute }}+W_{\text {solvent }} =2 \times V_{\text {solution }} .......(4)\end{aligned}$

We know that molality is given by

$m=\frac{n_{\text {solute }} \times 1000}{n_{\text {solvent }} \times W_{\text {solvent }}}$

Substituting from Eq. (3), we have

$m=\left(\frac{0.1}{0.9}\right) \times \frac{1000}{W_{\text {solvent }}}$ ......(5)

Molarity is given by

$M=\frac{n_{\text {solute }} \times 1000}{n_{\text {solvent }} \times V_{\text {solution }}}$

Substituting from Eq. (3), we have

$M=\left(\frac{0.1}{0.9}\right) \times \frac{1000}{V_{\text {solution }}}$ ..........(6)

Given that, Molarity $(M)=$ Molality $(m)$ ;

Therefore, from Eqs. (5) and (6), we get

$\begin{aligned}& \frac{0.1 \times 1000}{0.9 \times W_{\text {solvent }}} =\frac{0.1 \times 1000}{0.9 \times V_{\text {solution }}} \\\\& W_{\text {solvent }} =V_{\text {solution }}\end{aligned}$

From Eq. (4), we have

$\begin{aligned}W_{\text {solvent }} & =\frac{W_{\text {solute }}+W_{\text {solvent }}}{2} \\\\2 W_{\text {solvent }} & =W_{\text {solute }}+W_{\text {solvent }} \\\\W_{\text {solvent }} & =W_{\text {solute }} .........(7)\end{aligned}$

The molecular weight (MW) of the solute can be calculated by dividing the weight of the solute by the number of moles of solute. This can be written as :

$MW_{\text{solute}} = \frac{W_{\text{solute}}}{n_{\text{solute}}}$

2. Similarly, the molecular weight (MW) of the solvent can be calculated by dividing the weight of the solvent by the number of moles of solvent. This can be written as :

$MW_{\text{solvent}} = \frac{W_{\text{solvent}}}{n_{\text{solvent}}}$

In these formulas, $MW_{\text{solute}}$ and $MW_{\text{solvent}}$ represent the molecular weights of the solute and solvent respectively, $W_{\text{solute}}$ and $W_{\text{solvent}}$ represent their weights, and $n_{\text{solute}}$ and $n_{\text{solvent}}$ represent the number of moles of solute and solvent respectively.

From Eq. (3), we have

$\frac{n_{\text {solute }}}{n_{\text {solvent }}}=\frac{W_{\text {solute }} / M W_{\text {solute }}}{W_{\text {solvent }} / M W_{\text {solvent }}}=\frac{0.1}{0.9}$

Using Eq. (7), we get

$\frac{M W_{\text {solute }}}{M W_{\text {solvent }}}=9$