Solution A is prepared by dissolving 1 g of a protein (molar mass = 50,000 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 the same temperature. The osmotic pressure of resultant solution is z bar. x, y and z respectively are: (R = 0.083 L bar mol-1 K-1)
- 9.96 10-4; 9.96 10-4; 9.96 10-4
- 9.96 10-4; 9.96 10-4; 19.92 10-4
- 4.98 10-4; 4.98 10-4; 9.96 10-4
- 4.98 10-4; 4.98 10-4; 4.98 10-4
Answer
(1) 9.96 10-4; 9.96 10-4; 9.96 10-4
Solution
= CRT, so equal concentrations give equal osmotic pressures.
A: n = 150000 = 2 10-5 mol ⇒ C = 2 10-50.5 = 4 10-5 M
B: n = 250000 = 4 10-5 mol ⇒ C = 4 10-51 = 4 10-5 M
Mixture: n = 6 10-5 mol in 1.5 L ⇒ C = 4 10-5 M
⇒ x = y = z = 4 10-5 0.083 300
= 9.96 10-4 bar
Answer: (1)