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(A) The elevation in boiling point is given by the formula $\Delta T_b = i 	imes K_b 	imes m$. Since $K_b$ is constant for the solvent,$\Delta T_b$

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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 $\Delta T_b = i 	imes K_b 	imes m$. Since $K_b$ is constant for the solvent,$\Delta T_b$ depends on the product of the van't Hoff factor $(i)$ and molality $(m)$. For $0.01 \ M \ Na_2SO_4$,$i = 3$ (as it dissociates into $2Na^+ + SO_4^{2-}$),so $i 	imes m = 3 	imes 0.01 = 0.03$. For $0.01 \ M \ KNO_3$,$i = 2$ (as it dissociates into $K^+ + NO_3^-$),so $i 	imes m = 2 	imes 0.01 = 0.02$. For $0.015 \ M$ urea,$i = 1$ (non-electrolyte),so $i 	imes m = 1 	imes 0.015 = 0.015$. For $0.015 \ M$ 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 aqueous solutions will exhibit the highest boiling point?

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