JEE Main 2019 — Chemistry solution

JEE Main2019Chemistry

A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol^–1) and 1.8 g of glucose (molar mass = 180 g mol^–1) in 100 mL of water at 27°C. The osmotic pressure of the solution is

(R = 0.08206 L atm K^–1 mol^–1)

A
1.64 atm
B
2.46 atm
C
8.2 atm
D
4.92 atm

Solution

Osmotic pressure ( $π$ ) = CRT

Solute : urea and glucose

$∴$ $π$ = (C₁ + C₂) RT

$= (\frac{0 .6}{60 \times 0 .1} + \frac{1 .8}{180 \times 0 .1}) \times 0 .0821 \times 300$

$= 0 .2 \times 0 .0821 \times 300$

$= 4 .926 atm$

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