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(C) For dilute solutions,the relative lowering of vapour pressure is equal to the mole fraction of the solute,which is approximately equal to the rati

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$6$

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(C) For dilute solutions,the relative lowering of vapour pressure is equal to the mole fraction of the solute,which is approximately equal to the ratio of the number of moles of solute to the number of moles of solvent.$\frac{P_0 - P_s}{P_0} \approx \frac{n_{	ext{solute}}}{n_{	ext{solvent}}}$For the urea solution: $n_{	ext{urea}} = \frac{1 \ g}{60 \ g/mol} = \frac{1}{60} \ mol$,$n_{	ext{water}} = \frac{50 \ g}{18 \ g/mol} = \frac{50}{18} \ mol$.Relative lowering for urea $= \frac{1/60}{50/18} = \frac{1}{60} 	imes \frac{18}{50} = \frac{18}{3000} = \frac{3}{500}$.For the glucose solution: $n_{	ext{glucose}} = \frac{w}{180} \ mol$,$n_{	ext{water}} = \frac{100}{18} \ mol$.Relative lowering for glucose $= \frac{w/180}{100/18} = \frac{w}{180} 	imes \frac{18}{100} = \frac{w}{1000}$.Equating the two: $\frac{w}{1000} = \frac{3}{500} \implies w = \frac{3 	imes 1000}{500} = 6 \ g$.

The mass of glucose that should be dissolved in $100 \ g$ of water in order to produce the same relative lowering of vapour pressure as is produced by dissolving $1 \ g$ of urea (molar mass $= 60 \ g/mol$) in $50 \ g$ of water is ........... $g$. (Assume dilute solution in both cases):-

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