JEE Main20262 April 2026Evening ShiftChemistrySolutionsActual
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 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 ar
Options
- A9.96 10⁻⁴; 9.96 10⁻⁴; 9.96 10⁻⁴
- B9.96 10⁻⁴; 9.96 10⁻⁴; 19.92 10⁻⁴
- C4.98 10⁻⁴; 4.98 10⁻⁴; 9.96 10⁻⁴
- D4.98 10⁻⁴; 4.98 10⁻⁴; 4.98 10⁻⁴
Correct answer
A. 9.96 10⁻⁴; 9.96 10⁻⁴; 9.96 10⁻⁴
Step-by-step solution
Osmotic pressure is given by = CRT = w MV RT For solution A: x = 1 50000 0.5 0.083 300 x = 1 25000 24.9 = 9.96 10⁻⁴ bar For solution B: y = 2 50000 1 0.083 300 y = 2 50000 24.9 = 9.96 10⁻⁴ bar For the resultant solution (mixture of A and B): Total mass of protein = 1 + 2 = 3 g Total volume of solution = 0.5 + 1 = 1.5 L z = 3 50000 1.5 0.083 300 z = 3 75000 24.9 = 9.96 10⁻⁴ bar Therefore, x = y = z = 9.96 10⁻⁴ bar. Answer: 9.96 10⁻⁴; 9.96 10⁻⁴; 9.96 10⁻⁴