The osmotic pressure of a solution containing 40 g of solute (molecular mass 246) per litre at $27^{o} C$ is:
$[R = 0.0822 a t m L {m o l}^{- 1}]$

3.0 atm

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4.0 atm

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2.0 atm

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1.0 atm

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Solution

The correct option is B 4.0 atm
$π = C R T = \frac{(\frac{W_{2}}{M w_{2}}) R T}{V}$

Given $W_{2} = 40 g$

$M w_{2} = 246$

$T = 27^{o} C = 300 K$

V=1 L

Substituting all the values, we get

$π = \frac{\frac{40}{246} \times 0.082 \times 300}{1} = 4 a t m$

Hence, the correct option is $D$

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