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 are: (R = 0.083 L bar mol^ -1 K^ -1 )
A. 9.96 10^ -4 ; 9.96 10^ -4 ; 9.96 10^ -4
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
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