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(A) The elevation in boiling point is given by the formula $\Delta T_b = i 	imes K_b 	imes m$,where $i$ is the van't Hoff factor and $m$ is the mola

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$0.01 \ M \ Na_2SO_4$

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(A) The elevation in boiling point is given by the formula $\Delta T_b = i 	imes K_b 	imes m$,where $i$ is the van't Hoff factor and $m$ is the molality.Since $K_b$ is constant for a given solvent,the boiling point depends on the product $i 	imes m$.For $0.01 \ M \ Na_2SO_4$,$i = 3$ $(Na_2SO_4 ightarrow 2Na^+ + SO_4^{2-})$,so $i 	imes m = 3 	imes 0.01 = 0.03$.For $0.01 \ M \ KNO_3$,$i = 2$ $(KNO_3 ightarrow K^+ + NO_3^-)$,so $i 	imes m = 2 	imes 0.01 = 0.02$.For $0.015 \ M \ 	ext{urea}$,$i = 1$ (non-electrolyte),so $i 	imes m = 1 	imes 0.015 = 0.015$.For $0.015 \ M \ 	ext{glucose}$,$i = 1$ (non-electrolyte),so $i 	imes m = 1 	imes 0.015 = 0.015$.Since $0.01 \ M \ Na_2SO_4$ has the highest value of $i 	imes m$,it will exhibit the highest boiling point.

Which of the following solutions will exhibit the highest boiling point?

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