An aqueous solution containing one gram of urea boils at 100.25°C. The aqueous solution containing 3g of glucose in the same volume will boil at

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100°C

100.25°C

100.5°C

100.75°C

answer is B.

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Detailed Solution

$W_{u r e a} = 1 g, W_{g l u \cos e} = 3 g \frac{{(∆ T_{b})}_{urea}}{{(∆ T_{b})}_{glucose}} = \frac{n_{urea}}{n_{glucose}} = \frac{1}{60} \times \frac{180}{3} = 1 {(∆ T_{b})}_{urea} = {(∆ T_{b})}_{glucose} ∴ B . P of urea = B . P of glucose$

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