Question

‘x’ be the osmotic pressure of a solution formed by dissolving 1g of a protein (M = 50,000 g/mol) in 0.5 litre and ‘y’ be the osmotic pressure of solution formed by dissolving 2g of the same protein in 1 litre at 300 K. If ‘z’ be the osmotic pressure of solution formed by mixing above two solutions. Then the value of ‘x’, ‘y’ and ‘z’ respectively are.
\text{Use: \( R = 0.083 \, \text{lit-bar/K-mol} \)}

• A. \( 9.96 \times 10^{-4} \, \text{bar}, 9.96 \times 10^{-4} \, \text{bar}, 4.48 \times 10^{-4} \, \text{bar} \)
• B. \( 9.96 \times 10^{-4} \, \text{bar}, 19.2 \times 10^{-4} \, \text{bar}, 9.96 \times 10^{-4} \, \text{bar} \)
• C. \( 19.2 \times 10^{-4} \, \text{bar}, 19.2 \times 10^{-4} \, \text{bar}, 19.2 \times 10^{-4} \, \text{bar} \)
• D. \( 9.96 \times 10^{-4} \, \text{bar}, 9.96 \times 10^{-4} \, \text{bar}, 9.96 \times 10^{-4} \, \text{bar} \)

Hint:

Remember that the osmotic pressure is directly proportional to the number of moles of solute and inversely proportional to the volume of the solution.

Answer 1

The Correct Option is D

Step 1: Using the formula for osmotic pressure.
The osmotic pressure (\( \Pi \)) of a solution is calculated using the relation: \[ \Pi = \frac{nRT}{V} \] where: - \( n \) is the number of moles of solute, - \( R \) is the gas constant (\( 0.083 \, \text{L·bar·K}^{-1}\text{·mol}^{-1} \)), - \( T \) is the temperature (\( 300 \, \text{K} \)), - \( V \) is the volume of the solution in litres.
Step 2: Calculating the value of \( x \).
For 1 g of protein: - Molar mass \( M = 50{,}000 \, \text{g/mol} \) - Number of moles: \[ n = \frac{1}{50{,}000} = 2 \times 10^{-5} \, \text{mol} \] - Volume \( V = 0.5 \, \text{L} \) Substituting into the formula: \[ \Pi = \frac{(2 \times 10^{-5})(0.083)(300)}{0.5} \] \[ \Pi = 9.96 \times 10^{-4} \, \text{bar} \] Thus, \[ x = 9.96 \times 10^{-4} \, \text{bar} \]
Step 3: Calculating the value of \( y \).
For 2 g of protein: - Number of moles: \[ n = \frac{2}{50{,}000} = 4 \times 10^{-5} \, \text{mol} \] - Volume \( V = 1 \, \text{L} \) Substituting into the formula: \[ \Pi = \frac{(4 \times 10^{-5})(0.083)(300)}{1} \] \[ \Pi = 9.96 \times 10^{-4} \, \text{bar} \] Hence, \[ y = 9.96 \times 10^{-4} \, \text{bar} \]
Step 4: Calculating the value of \( z \).
When the two solutions are mixed, the total osmotic pressure is the sum of the individual osmotic pressures: \[ z = x + y \] \[ z = 9.96 \times 10^{-4} + 9.96 \times 10^{-4} \] \[ z = 1.992 \times 10^{-3} \, \text{bar} \]
Step 5: Conclusion.
Therefore, the osmotic pressures are: \[ x = 9.96 \times 10^{-4} \, \text{bar}, \quad y = 9.96 \times 10^{-4} \, \text{bar}, \quad z = 1.992 \times 10^{-3} \, \text{bar} \]
Final Answer: \( x = 9.96 \times 10^{-4} \, \text{bar}, \; y = 9.96 \times 10^{-4} \, \text{bar}, \; z = 1.992 \times 10^{-3} \, \text{bar} \).