When 1 g of urea is dissolved in 50 g of wa | vedclass

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(B) The relative lowering of vapour pressure is given by the formula: $\frac{p^0 - p}{p^0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{W \

Vedclass · Accepted Answer

$3$

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(B) The relative lowering of vapour pressure is given by the formula: $\frac{p^0 - p}{p^0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{W 	imes M_0}{M 	imes W_0}$.For urea $(M = 60 \ g/mol)$: $\frac{p^0 - p}{p^0} = \frac{1 	imes 18}{60 	imes 50}$.For glucose $(M = 180 \ g/mol)$: $\frac{p^0 - p}{p^0} = \frac{W 	imes 18}{180 	imes 50}$.Since the relative lowering of vapour pressure is equal for both:$\frac{1 	imes 18}{60 	imes 50} = \frac{W 	imes 18}{180 	imes 50}$.Solving for $W$:$\frac{1}{60} = \frac{W}{180} \implies W = \frac{180}{60} = 3 \ g$.

When $1 \ g$ of urea is dissolved in $50 \ g$ of water,the relative lowering of vapour pressure of the solution is equal to the relative lowering of vapour pressure of a solution formed by dissolving glucose in the same amount of water. How many grams of glucose were dissolved?

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