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What will be the mass of glucose in grams that should be dissolved in 50 g of water in order to produce the same lowering of vapour pressure by dissolving 1 g of urea in the same quantity of water?

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answer is 3.

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Detailed Solution

$\frac{p^{o} {-p}_{}}{p^{o}} = \frac{n_{2}}{n_{1}} = \frac{W_{2} M_{2}}{W_{1} M_{2}}$
As $\frac{p^{o} {-p}_{}}{p^{o}} = \frac{n_{2}}{n_{1}} same in the two cases.$
${(\frac{W_{3} M_{2}}{W_{1} M_{2}})}_{glucose} = {(\frac{W_{2} M_{1}}{W_{1} M_{2}})}_{urea} \Rightarrow \frac{W_{2} \times 18}{50 \times 180} = \frac{1 \times 18}{50 \times 60} \Rightarrow W_{2} = 3 g$

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