JEE Advance - Chemistry (2016 - Paper 1 Offline - No. 3)
The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its
molality. Density of this solution at 298 K is 2.0 g cm–3 . The ratio of the molecular weights of the solute and solvent, $$\left( {{{M{W_{solute}}} \over {M{W_{solvent}}}}} \right)$$, is
Answer
9
Explanation
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
$$ \frac{M W_{\text {solute }}}{M W_{\text {solvent }}}=9 $$
$$ x_{\text {solute }}=\frac{n_{\text {solute }}}{\left(n_{\text {solute }}+n_{\text {solvent }}\right)}=0.1 $$ ......(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
$$ \frac{M W_{\text {solute }}}{M W_{\text {solvent }}}=9 $$
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