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← JEE Advanced 2023 · Paper 2
ChemistryInteger+4Question 46 of 51

JEE Advanced 2023 · Paper 2, Chemistry question 12

50 mL of 0.2 molal urea solution (density =1.012 g mL−1= 1.012\ \mathrm{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= 60\ \mathrm{g\,mol^{-1}}; gas constant, R=62 L Torr K−1 mol−1R = 62\ \mathrm{L\,Torr\,K^{-1}\,mol^{-1}}; Assume, ΔmixH=0\Delta_\mathrm{mix}H = 0, ΔmixV=0\Delta_\mathrm{mix}V = 0]

Answer and worked solution

Official answer: 682

First solution. 50 mL has mass 50.650.6 g. With solvent mass WW (g), the urea is 0.2×W10000.2\times\frac{W}{1000} mol, weighing 0.012W0.012W g. So
W+0.012W=50.6  ⟹  W=50 g.W+0.012W=50.6\implies W=50\ \text{g}.
That is 0.010.01 mol of urea.

Second solution. 0.060.06 g of urea is 0.0010.001 mol.

After mixing. There are 0.011 mol in 300 mL (no volume change on mixing):
π=CRT=0.0110.3×62×300=682 Torr.\pi=CRT=\frac{0.011}{0.3}\times62\times300=682\ \text{Torr}.

Topics:Colligative propertiesConcentration and solubility
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