What would the boiling point elevation of 0.1 | vedclass

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(B) The boiling point elevation is given by the formula $\Delta T_b = i 	imes K_b 	imes m$. Since $K_b$ and $m$ are constant for the given solutions

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$0.1 \ m$ sodium chloride

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(B) The boiling point elevation is given by the formula $\Delta T_b = i 	imes K_b 	imes m$. Since $K_b$ and $m$ are constant for the given solutions,the boiling point elevation depends on the van't Hoff factor $(i)$.For $KNO_3$,it dissociates as $K^+ + NO_3^-$,so $i = 2$.For $0.1 \ m$ Urea,$i = 1$ (non-electrolyte).For $0.1 \ m$ sodium chloride $(NaCl)$,it dissociates as $Na^+ + Cl^-$,so $i = 2$.For $0.1 \ m$ potassium sulfate $(K_2SO_4)$,it dissociates as $2K^+ + SO_4^{2-}$,so $i = 3$.For $0.1 \ m$ aluminum nitrate $(Al(NO_3)_3)$,it dissociates as $Al^{3+} + 3NO_3^-$,so $i = 4$.Therefore,the boiling point elevation of $0.1 \ m$ $KNO_3$ is equal to that of $0.1 \ m$ sodium chloride.

What would the boiling point elevation of $0.1 \ m$ $KNO_3$ be equal to?

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