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Osmotic Pressure

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Question

A solution of urea contains 8.6 g per litre. It is isotonic with $5 %$ solution of a non-volatile solute. The molecular mass of the solute will be: (in amu)

A

349

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B

34.9

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C

3490

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D

861

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Solution

The correct option is A 349
No of moles of urea = $\frac{8.6}{60}$

5% of unknown solution $= \frac{5 g}{100 m l}$

$\frac{50 g}{1000 m l}$

$= 5 g / l i t r e$

No of moles of unknown solute $= \frac{50}{m}$

$\frac{8.6}{60}$ = $= \frac{50}{m}$

$m = 348.9$ ~ $349 a m u$

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