JEE Advanced 2023 — Paper 2 — CHEM-III solution

JEE Advanced2023Paper 2CHEM-III

$50$ 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, $Δ_{mi x} H = 0$ , $Δ_{mi x} V = 0$ ]

Your answer

Solution

Step 1. Urea in the first solution. A $0.2$ molal solution contains $0.2$ mol $= 12$ g of urea per $1000$ g of solvent, so the total mass of $1$ kg of solvent + $12$ g urea $= 1012$ g of solution carries $12$ g of urea.

Mass of $50$ mL of this solution $= 50 \times 1.012 = 50.6$ g.

Mass of urea in this aliquot $= 12 \times \frac{50.6}{1012} = 0.6$ g.

Step 2. Total urea after mixing. Adding the $0.06$ g of urea from the second solution gives $0.66$ g of urea, that is, $\frac{0.66}{60} = 0.011$ mol.

Step 3. Total volume. $50 + 250 = 300$ mL $= 0.3$ L (volumes are additive since $Δ_{mi x} V = 0$ ).

Step 4. Osmotic pressure.

$π = \frac{n}{V} R T = \frac{0.011}{0.3} \times 62 \times 300 = 682 Torr$ .

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