5 text{ g} of urea is dissolved in 100 text{ | vedclass

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(C) Step $1$: Calculate the molality of the urea solution.Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}}$.Moles of urea $

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$18$

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(C) Step $1$: Calculate the molality of the urea solution.Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}}$.Moles of urea $= \frac{5 	ext{ g}}{60 	ext{ g mol}^{-1}} = \frac{1}{12} 	ext{ mol}$.Molality of urea solution $(m_1) = \frac{1/12 	ext{ mol}}{0.1 	ext{ kg}} = \frac{1}{1.2} 	ext{ mol kg}^{-1}$.Step $2$: Since boiling points are the same, the molality of the glucose solution must be equal to the molality of the urea solution $(m_2 = m_1)$.Let $x$ be the mass of glucose.Moles of glucose $= \frac{x}{180} 	ext{ mol}$.Molality of glucose solution $(m_2) = \frac{x/180 	ext{ mol}}{0.12 	ext{ kg}} = \frac{x}{180 	imes 0.12} 	ext{ mol kg}^{-1} = \frac{x}{21.6} 	ext{ mol kg}^{-1}$.Step $3$: Equate the molalities:$\frac{1}{1.2} = \frac{x}{21.6}$.$x = \frac{21.6}{1.2} = 18 	ext{ g}$.Therefore, the required amount of glucose is $18 	ext{ g}$.

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