50   mL of 0 . 2 molal urea solution (density = 1 . 012   g   mL - 1 at 300   K ) is mixed with 250   mL of a solution containing 0 . 06   g of urea. Both the solutions were prepared in the same solvent. The osmotic pressure (in torr) of the resulting at 300   K is [Use : Molar mass of urea = 60   g   mol - 1 ; gas constant, R = 62   L - torr   K - 1   mol - 1 ; Assume, Δ mix H = 0 , Δ mix  V = 0
A. A
Mole of urea = 0 . 2 Weight of urea = moles of urea × molar mass Weight of urea = 0 . 2 × 60 = 12   g Weight of solvent = 1000   g Weight of solution = 1012   g Volume of solution = Weight of solution density of solution ∴ Volume of solution = 1012 1 . 012 = 1000   ml ∵    1000   ml solution contain 0 . 2 mole ∴ 50   ml solution contain = 0 . 2 × 50 1000 = 0 . 01 Mole of urea in other solution = 0 . 06 60 = 0 . 001 ∴ Concentration of solution = 0 . 01 + 0 . 001 300 1000 Now, osmotic pressure can be calculated as follows, ∴   π = CRT = 0 . 0366 × 62 × 300 = 682
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