A solution has an osmotic pressure of 'x' kPa at $300 \mathrm{~K}$ having 1 mole of solute in $10 \cdot 5…

A solution has an osmotic pressure of 'x' kPa at $300 \mathrm{~K}$ having 1 mole of solute in $10 \cdot 5 \mathrm{~m}^{3}$ of solution. If it's osmotic pressure is reduced to $\left(\frac{1}{10}\right)^{\text {th }}$ of it's initial value, what is the new volume of solution?
  1. $30 \mathrm{~m}^{3}$
  2. $105 \mathrm{~m}^{3}$
  3. $110 \mathrm{~m}^{3}$
  4. $11.0 \mathrm{~m}^{3}$

Solution

$\begin{array}{l} \pi=\mathrm{X} \mathrm{kPa}, \quad \mathrm{V}=10.5 \mathrm{~m}^{3}, \pi^{\prime}=\frac{\mathrm{X}}{10} \mathrm{kPa}, \quad \mathrm{V}^{\prime}=? \\ \pi=\frac{\mathrm{n}}{\mathrm{v}} \mathrm{RT} \end{array}$ i. $X=\frac{1}{10.5} \times 8.314 \times 300$; ii. $\frac{X}{10}=\frac{1}{V^{\prime}} \times 8.314 \times 300$ By dividing equation (i) by (ii) $\begin{aligned} \frac{X}{X / 10} &=\frac{\frac{1}{10.5} \times 8.314 \times 300}{\frac{1}{1 / V^{\prime} \times 8.314 \times 300}} \\ 10 &=\frac{V^{\prime}}{10.5} \quad \therefore V^{\prime}=105 \mathrm{~m}^{3} \end{aligned}$

Asked in: MHT CET 2020 (16 Oct Shift 1)

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