Question:
Solution A is prepared by dissolving 1 g of a protein (molar mass = 50000 g mol\(^{-1}\)) in 0.5 L of water at 300 K. Its osmotic pressure is \( x \) bar. Solution B is made by dissolving 2 g of the same protein in 1 L of water at 300 K. Osmotic pressure of solution B is \( y \) bar. Entire solution of A is mixed with entire solution of B at same temperature. The osmotic pressure of resultant solution is \( z \) bar. \( x \), \( y \), and \( z \) respectively are:
Solution A is prepared by dissolving 1 g of a protein (molar mass = 50000 g mol\(^{-1}\)) in 0.5 L of water at 300 K. Its osmotic pressure is \( x \) bar. Solution B is made by dissolving 2 g of the same protein in 1 L of water at 300 K. Osmotic pressure of solution B is \( y \) bar. Entire solution of A is mixed with entire solution of B at same temperature. The osmotic pressure of resultant solution is \( z \) bar. \( x \), \( y \), and \( z \) respectively are:
Updated On: Apr 12, 2026
- \( 9.96 \times 10^{-4}, 9.96 \times 10^{-4}, 9.96 \times 10^{-4} \)
- \( 9.96 \times 10^{-4}, 9.96 \times 10^{-4}, 19.92 \times 10^{-4} \)
- \( 9.96 \times 10^{-4}, 4.98 \times 10^{-4}, 9.96 \times 10^{-4} \)
- \( 4.98 \times 10^{-4}, 4.98 \times 10^{-4}, 4.98 \times 10^{-4} \)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding osmotic pressure.
The osmotic pressure \( \Pi \) of a solution is given by the formula: \[ \Pi = \frac{nRT}{V} \] where \( n \) is the number of moles of solute, \( R \) is the universal gas constant, \( T \) is the temperature, and \( V \) is the volume of the solution. Osmotic pressure is directly proportional to the number of moles of solute per volume.
Step 2: Calculating the osmotic pressure for solution A.
For solution A, the number of moles \( n_A \) of protein is: \[ n_A = \frac{1}{50000} = 2 \times 10^{-5} \, \text{mol} \] The osmotic pressure of solution A is: \[ \Pi_A = \frac{n_A RT}{V_A} = \frac{2 \times 10^{-5} \times 0.083 \times 300}{0.5} = 9.96 \times 10^{-4} \, \text{bar} \] Thus, \( x = 9.96 \times 10^{-4} \).
Step 3: Calculating the osmotic pressure for solution B.
For solution B, the number of moles \( n_B \) of protein is: \[ n_B = \frac{2}{50000} = 4 \times 10^{-5} \, \text{mol} \] The osmotic pressure of solution B is: \[ \Pi_B = \frac{n_B RT}{V_B} = \frac{4 \times 10^{-5} \times 0.083 \times 300}{1} = 9.96 \times 10^{-4} \, \text{bar} \] Thus, \( y = 9.96 \times 10^{-4} \).
Step 4: Calculating the osmotic pressure of the resultant solution.
When solutions A and B are mixed, the total number of moles is \( n_A + n_B = 6 \times 10^{-5} \). The total volume is \( V_A + V_B = 1.5 \, \text{L} \). The osmotic pressure of the resultant solution is: \[ \Pi_{\text{total}} = \frac{(n_A + n_B)RT}{V_A + V_B} = \frac{6 \times 10^{-5} \times 0.083 \times 300}{1.5} = 19.92 \times 10^{-4} \, \text{bar} \] Thus, \( z = 19.92 \times 10^{-4} \).
Final Answer: (B) \( 9.96 \times 10^{-4}, 9.96 \times 10^{-4}, 19.92 \times 10^{-4} \)
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