The osmotic pressure exerted by a solution prepared by dissolving 2.0 g of protein of molar mass $60 \mathrm{~kg} \mathrm{~mol}^{-1}$ in 200 mL of water at $27^{\circ} \mathrm{C}$ is $\_\_\_\_$ Pa. [integer value] (use $\mathrm{R}=0.083 \mathrm{~L}$ bar $\mathrm{mol}^{-1} \mathrm{~K}^{-1}$)

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Solutions

The osmotic pressure exerted by a solution prepared by dissolving 2.0 g of protein of molar mass

60 kg mol^{- 1}

in 200 mL of water at

2 7^{\circ} C

____

Pa. [integer value]
(use

R = 0.083 L

bar

mol^{- 1} K^{- 1}

)

Answer: 415

Solution

Using the osmotic pressure formula

π = i C R T

where

i = 1

for a non-electrolyte (protein).

First, calculate the concentration:

C = \frac{n}{V} = \frac{2/ ( 60000 )}{0.2} = \frac{2}{12000}

mol/L.

Then:

π = 1 \times \frac{2}{12000} \times 0.083 \times 300 = 0.00415

bar.

Converting to Pa:

0.00415 \times 1 0^{5} = 415

Pa.

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Integer answer

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