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: \( R = 0.083 \, \text{lit-bar/K-mol} \)}
‘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: \( R = 0.083 \, \text{lit-bar/K-mol} \)}
\text{Use: \( R = 0.083 \, \text{lit-bar/K-mol} \)}
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Remember that the osmotic pressure is directly proportional to the number of moles of solute and inversely proportional to the volume of the solution.
Updated On: Apr 4, 2026
- \( 9.96 \times 10^{-4} \, \text{bar}, 9.96 \times 10^{-4} \, \text{bar}, 4.48 \times 10^{-4} \, \text{bar} \)
- \( 9.96 \times 10^{-4} \, \text{bar}, 19.2 \times 10^{-4} \, \text{bar}, 9.96 \times 10^{-4} \, \text{bar} \)
- \( 19.2 \times 10^{-4} \, \text{bar}, 19.2 \times 10^{-4} \, \text{bar}, 19.2 \times 10^{-4} \, \text{bar} \)
- \( 9.96 \times 10^{-4} \, \text{bar}, 9.96 \times 10^{-4} \, \text{bar}, 9.96 \times 10^{-4} \, \text{bar} \)
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The Correct Option is D
Solution and Explanation
Step 1: Use the formula for osmotic pressure.
The osmotic pressure (\( \Pi \)) is given by: \[ \Pi = \frac{nRT}{V} \] Where: - \( n \) is the number of moles of the solute, - \( R \) is the gas constant (\( 0.083 \, \text{lit-bar/K-mol} \)), - \( T \) is the temperature in Kelvin (\( 300 \, \text{K} \)), - \( V \) is the volume in litres.
Step 2: Calculate ‘x’.
For 1g of protein: - Molar mass of protein \( M = 50,000 \, \text{g/mol} \), - Number of moles \( n = \frac{1}{50,000} = 2 \times 10^{-5} \, \text{mol} \), - Volume \( V = 0.5 \, \text{litres} \). Substituting into the osmotic pressure formula: \[ \Pi = \frac{(2 \times 10^{-5})(0.083)(300)}{0.5} = 9.96 \times 10^{-4} \, \text{bar} \] So, \( x = 9.96 \times 10^{-4} \, \text{bar} \).
Step 3: Calculate ‘y’.
For 2g of protein: - Moles \( n = \frac{2}{50,000} = 4 \times 10^{-5} \, \text{mol} \), - Volume \( V = 1 \, \text{litre} \). Substituting into the osmotic pressure formula: \[ \Pi = \frac{(4 \times 10^{-5})(0.083)(300)}{1} = 1.98 \times 10^{-3} \, \text{bar} \] So, \( y = 1.98 \times 10^{-3} \, \text{bar} \).
Step 4: Calculate ‘z’.
The osmotic pressure when mixing both solutions is the sum of the individual pressures: \[ z = x + y = 9.96 \times 10^{-4} + 1.98 \times 10^{-3} = 2.98 \times 10^{-3} \, \text{bar} \]
Step 5: Conclusion.
Therefore, the values of \( x \), \( y \), and \( z \) are: \[ x = 9.96 \times 10^{-4} \, \text{bar}, \quad y = 9.96 \times 10^{-4} \, \text{bar}, \quad z = 9.96 \times 10^{-4} \, \text{bar} \] Final Answer: \( x = 9.96 \times 10^{-4} \, \text{bar}, y = 9.96 \times 10^{-4} \, \text{bar}, z = 9.96 \times 10^{-4} \, \text{bar} \)
The osmotic pressure (\( \Pi \)) is given by: \[ \Pi = \frac{nRT}{V} \] Where: - \( n \) is the number of moles of the solute, - \( R \) is the gas constant (\( 0.083 \, \text{lit-bar/K-mol} \)), - \( T \) is the temperature in Kelvin (\( 300 \, \text{K} \)), - \( V \) is the volume in litres.
Step 2: Calculate ‘x’.
For 1g of protein: - Molar mass of protein \( M = 50,000 \, \text{g/mol} \), - Number of moles \( n = \frac{1}{50,000} = 2 \times 10^{-5} \, \text{mol} \), - Volume \( V = 0.5 \, \text{litres} \). Substituting into the osmotic pressure formula: \[ \Pi = \frac{(2 \times 10^{-5})(0.083)(300)}{0.5} = 9.96 \times 10^{-4} \, \text{bar} \] So, \( x = 9.96 \times 10^{-4} \, \text{bar} \).
Step 3: Calculate ‘y’.
For 2g of protein: - Moles \( n = \frac{2}{50,000} = 4 \times 10^{-5} \, \text{mol} \), - Volume \( V = 1 \, \text{litre} \). Substituting into the osmotic pressure formula: \[ \Pi = \frac{(4 \times 10^{-5})(0.083)(300)}{1} = 1.98 \times 10^{-3} \, \text{bar} \] So, \( y = 1.98 \times 10^{-3} \, \text{bar} \).
Step 4: Calculate ‘z’.
The osmotic pressure when mixing both solutions is the sum of the individual pressures: \[ z = x + y = 9.96 \times 10^{-4} + 1.98 \times 10^{-3} = 2.98 \times 10^{-3} \, \text{bar} \]
Step 5: Conclusion.
Therefore, the values of \( x \), \( y \), and \( z \) are: \[ x = 9.96 \times 10^{-4} \, \text{bar}, \quad y = 9.96 \times 10^{-4} \, \text{bar}, \quad z = 9.96 \times 10^{-4} \, \text{bar} \] Final Answer: \( x = 9.96 \times 10^{-4} \, \text{bar}, y = 9.96 \times 10^{-4} \, \text{bar}, z = 9.96 \times 10^{-4} \, \text{bar} \)
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