JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Solutions

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^{- 1} mol^{- 1}

; Assume,

Δ_{mix} H = 0

Δ_{mix} V = 0]

Answer: 682

Solution

First solution: 50 mL of 0.2 molal urea

Mass of solution = $50 \times 1.012 = 50.6$ g
0.2 molal means 0.2 moles urea per 1000 g solvent
Mass of solution with 0.2 moles urea = $1000 + 0.2 \times 60 = 1012$ g
Mass of urea in 50.6 g solution = $\frac{0.2 \times 60 \times 50.6}{1012} = 0.6$ g

Second solution: 250 mL containing 0.06 g urea

Total urea =

0.6 + 0.06 = 0.66

g =

\frac{0.66}{60} = 0.011

moles
Total volume =

50 + 250 = 300

mL = 0.3 L

Concentration

C = \frac{0.011}{0.3}

mol/L

Osmotic pressure:

π = C R T = \frac{0.011}{0.3} \times 62 \times 300 = \frac{0.66 \times 62 \times 300}{60 \times 0.3} = 682 Torr

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Integer answer

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JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Solutions

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