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 osmo

Question

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

TestHub · Accepted Answer

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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