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:

• A. \( 9.96 \times 10^{-4}, 9.96 \times 10^{-4}, 9.96 \times 10^{-4} \)
• B. \( 9.96 \times 10^{-4}, 9.96 \times 10^{-4}, 19.92 \times 10^{-4} \)
• C. \( 9.96 \times 10^{-4}, 4.98 \times 10^{-4}, 9.96 \times 10^{-4} \)
• D. \( 4.98 \times 10^{-4}, 4.98 \times 10^{-4}, 4.98 \times 10^{-4} \)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Osmotic pressure depends directly on the molar concentration (Molarity) of the solution.
If two solutions have the identical concentration and temperature, they will have identical osmotic pressures.
Furthermore, mixing two solutions of the exact same concentration will result in a final mixture with that same constant concentration.
Step 2: Key Formula or Approach:
Molarity \(C = \frac{\text{moles of solute}}{\text{Volume of solution in L}} = \frac{\text{mass} / \text{Molar mass}}{V}\).
Osmotic pressure \(\pi = CRT\).
Step 3: Detailed Explanation:
Analyze Solution A:
Mass \(m_A = 1 \text{ g}\), Volume \(V_A = 0.5 \text{ L}\).
Concentration \(C_A = \frac{1 / 50000}{0.5} = \frac{2}{50000} = 4 \times 10^{-5} \text{ mol/L}\).
Calculate the osmotic pressure \(x\):
\[ x = \pi_A = C_A R T = (4 \times 10^{-5}) \times 0.083 \times 300 \] \[ x = 1200 \times 10^{-5} \times 0.083 = 99.6 \times 10^{-5} = 9.96 \times 10^{-4} \text{ bar} \] Analyze Solution B:
Mass \(m_B = 2 \text{ g}\), Volume \(V_B = 1 \text{ L}\).
Concentration \(C_B = \frac{2 / 50000}{1} = \frac{2}{50000} = 4 \times 10^{-5} \text{ mol/L}\).
Since \(C_A = C_B\) and temperature is the same, the osmotic pressure is identical:
\[ y = \pi_B = 9.96 \times 10^{-4} \text{ bar} \] Analyze the Mixed Solution:
Total mass = \(1 \text{ g} + 2 \text{ g} = 3 \text{ g}\).
Total volume = \(0.5 \text{ L} + 1 \text{ L} = 1.5 \text{ L}\).
Concentration \(C_{mix} = \frac{3 / 50000}{1.5} = \frac{2}{50000} = 4 \times 10^{-5} \text{ mol/L}\).
Since the final concentration remains identical, the final osmotic pressure \(z\) is also identical.
\[ z = \pi_{mix} = 9.96 \times 10^{-4} \text{ bar} \] Step 4: Final Answer:
The values are \(9.96 \times 10^{-4}; 9.96 \times 10^{-4}; 9.96 \times 10^{-4}\).