Question

A solution has an osmotic pressure of \( x \) kPa at 300K having 1 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. 30 m\(^3\)
• B. 105 m\(^3\)
• C. 110 m\(^3\)
• D. 11.0 m\(^3\)

Hint:

For osmotic pressure, remember that pressure is inversely proportional to volume. If pressure decreases, volume increases.

Answer 1

The Correct Option is B

Step 1: Understanding the relation.
The osmotic pressure \(\pi\) is related to the volume \(V\) of the solution and temperature \(T\) by the formula: \[ \pi = \frac{nRT}{V} \] Where: - \(\pi\) is the osmotic pressure, - \(n\) is the number of moles of solute, - \(R\) is the gas constant, - \(T\) is the temperature in Kelvin, - \(V\) is the volume of the solution.

Step 2: Initial and final osmotic pressures.
Initial osmotic pressure: \(\pi_1 = \frac{nRT}{V_1}\) Final osmotic pressure: \(\pi_2 = \frac{nRT}{V_2}\) Since the osmotic pressure is reduced to \( \frac{1}{10} \), we have: \[ \frac{\pi_2}{\pi_1} = \frac{1}{10} \] \[ \frac{V_1}{V_2} = \frac{1}{10} \] Thus: \[ V_2 = 10 \times V_1 = 10 \times 10^{-5} = 105 \, \text{m}^3 \]
Step 3: Conclusion.
The new volume of solution is 105 m\(^3\).