The osmotic pressure of a solution (density is $1 g {m L}^{- 1}$ ) containing $3 g$ of glucose (molecular weight $= 180$ ) in $60 g$ of water at $15^{o} C$ is:

0.34 a t m

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0.65 a t m

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6.25 a t m

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5.57 a t m

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Solution

The correct option is B $6.25 a t m$
$π \times \frac{63}{1000} = \frac{3}{180} \times 0.0821 \times 288$
$∴ π = 6.25 a t m$
$∵ V_{s o l u t i o n} =$ mass of solution $\times$ Density

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Similar questions

1 g m L^{- 1}

15^{\circ} C

15^{o} C

If 3 gm of glucose (mol. wt. 180) is dissolved in 60 gm of water at 15⁰C. Then the osmotic pressure of this solution will be

5 %

25^{o} C

= 180

R = 0.0821

K^{- 1} m o l^{- 1}

25^{o} C

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