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If osmotic pressure of $4 \%(\mathrm{w} / \mathrm{v})$ solution of sucrose is same as $2 \%(\mathrm{w} / \mathrm{v})$ solution of ' X ', then the molecular mass of $X(\mathrm{~g} / \mathrm{mol})$ is
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$171$
$4 \%$ solution of sucrose $(\mathrm{w} / \mathrm{v})$ means 4 g of sucrose is dissolved in 100 mL of solution. Similarly, $2 \%(\mathrm{w} / \mathrm{v}$ ) solution of ' X ' means 2 g of X present in 100 mL of solution.
As we know that, osmotic pressure $(\pi)=$ CRT or $\pi=\frac{n}{V} R T$; where $n=$ number of moles of solute.
Since, $\pi_1$ (for sucrose solution) $=\pi_2$
(for X solution)
$\therefore \quad \frac{\mathrm{n}_1}{\mathrm{~V}_1} \mathrm{RT}=\frac{\mathrm{n}_2}{\mathrm{~V}_2} \mathrm{RT}$
$\begin{aligned} & \Rightarrow \quad \frac{4}{342}=\frac{2}{\text { Molecular weight }} \\ & {\left[\because \mathrm{n}=\frac{\text { Mass of solute }}{\text { Molar mass of solute }}\right]} \\ & \Rightarrow \text { Molecular weight }=\frac{2}{4} \times 342\end{aligned}$
Molecular weight $=171 \mathrm{~g} \mathrm{~mol}^{-1}$
As we know that, osmotic pressure $(\pi)=$ CRT or $\pi=\frac{n}{V} R T$; where $n=$ number of moles of solute.
Since, $\pi_1$ (for sucrose solution) $=\pi_2$
(for X solution)
$\therefore \quad \frac{\mathrm{n}_1}{\mathrm{~V}_1} \mathrm{RT}=\frac{\mathrm{n}_2}{\mathrm{~V}_2} \mathrm{RT}$
$\begin{aligned} & \Rightarrow \quad \frac{4}{342}=\frac{2}{\text { Molecular weight }} \\ & {\left[\because \mathrm{n}=\frac{\text { Mass of solute }}{\text { Molar mass of solute }}\right]} \\ & \Rightarrow \text { Molecular weight }=\frac{2}{4} \times 342\end{aligned}$
Molecular weight $=171 \mathrm{~g} \mathrm{~mol}^{-1}$
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