If 3 g of glucose (molecular mass 180) is dissolved in 60 g of water at $15^{o} C$ , then the osmotic pressure of this solution will be :

0.34 atm

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0.65 atm

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6.57 atm

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5.57 atm

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Solution

The correct option is C 6.57 atm
$π = C R T$

$= \frac{3 \times 1000}{180 \times 60} \times 0.0821 \times 288$

$= 6.56 a t m$

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