Question:
A solution has an osmotic pressure of X kPa at 300 K having one mole of solute in \(10.5\ \text{m}^3\) of solution. If its osmotic pressure is reduced to \(\frac{1}{10}\) of its initial value, what is the new volume of solution?
A solution has an osmotic pressure of X kPa at 300 K having one mole of solute in \(10.5\ \text{m}^3\) of solution. If its osmotic pressure is reduced to \(\frac{1}{10}\) of its initial value, what is the new volume of solution?
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For dilute solutions at constant temperature, osmotic pressure varies inversely with volume.
Updated On: Feb 11, 2026
- 30 m\(^3\)
- 110 m\(^3\)
- 11.0 m\(^3\)
- 105 m\(^3\)
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The Correct Option is D
Solution and Explanation
Step 1: Write the formula for osmotic pressure.
\[ \pi = \frac{nRT}{V} \]
Step 2: Identify constants.
Number of moles \(n\) and temperature \(T\) remain constant.
Thus, osmotic pressure is inversely proportional to volume.
Step 3: Apply proportionality.
\[ \frac{\pi_1}{\pi_2} = \frac{V_2}{V_1} \]
Given \(\pi_2 = \frac{1}{10}\pi_1\).
Step 4: Calculate new volume.
\[ \frac{\pi_1}{\pi_1/10} = \frac{V_2}{10.5} \Rightarrow 10 = \frac{V_2}{10.5} \]
\[ V_2 = 105\ \text{m}^3 \]
Step 5: Conclusion.
The new volume of solution is 105 m\(^3\).
\[ \pi = \frac{nRT}{V} \]
Step 2: Identify constants.
Number of moles \(n\) and temperature \(T\) remain constant.
Thus, osmotic pressure is inversely proportional to volume.
Step 3: Apply proportionality.
\[ \frac{\pi_1}{\pi_2} = \frac{V_2}{V_1} \]
Given \(\pi_2 = \frac{1}{10}\pi_1\).
Step 4: Calculate new volume.
\[ \frac{\pi_1}{\pi_1/10} = \frac{V_2}{10.5} \Rightarrow 10 = \frac{V_2}{10.5} \]
\[ V_2 = 105\ \text{m}^3 \]
Step 5: Conclusion.
The new volume of solution is 105 m\(^3\).
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