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 solution at 300 K is__. [Use : Molar mass of urea = 60 g mol -1 ; gas constant, R = 62 L Torr K -1 mol -1 ; Assume, 
mix H = 0, A mix V= 0]
Text Solution
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Weight of 50 ml 0.2 molal urea = V x d = 50 x 1.012 = 50.6 gm
Given 0.2 molal implies
1000 gm solvent has 0.2 moles urea
So weight of solution = 1000 + 0.2 x 60 = 1012 gm.
So wt. of urea in 50.6 gm solution = 
= 0.6 gm
Total urea = 0.6 + 0.06 = 0.66 gm Total volume = 300 ml
Now, osmotic pressure 
= 682 Torr.
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