An aqueous solution containing 1 g of urea b | vedclass

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(C) The elevation in boiling point is given by $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.For urea: $m_1 = \frac{1 \ g / 60

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$100.25$

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(C) The elevation in boiling point is given by $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.For urea: $m_1 = \frac{1 \ g / 60 \ g \cdot mol^{-1}}{V \ L} = \frac{1}{60V} \ mol \cdot L^{-1}$.For glucose: $m_2 = \frac{3 \ g / 180 \ g \cdot mol^{-1}}{V \ L} = \frac{1}{60V} \ mol \cdot L^{-1}$.Since the molality $(m)$ is the same for both solutions,the elevation in boiling point $(\Delta T_b)$ will be identical.Given $\Delta T_b = 100.25 \ ^oC - 100 \ ^oC = 0.25 \ ^oC$ for urea.Therefore,the boiling point of the glucose solution will be $100 \ ^oC + 0.25 \ ^oC = 100.25 \ ^oC$.

An aqueous solution containing $1 \ g$ of urea boils at $100.25 \ ^oC$. The aqueous solution containing $3 \ g$ of glucose in the same volume will boil at ........ $^oC$. (Molecular weight of urea and glucose are $60$ and $180$ respectively)

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