The boiling point of 0.001 M aqueous solutio | vedclass

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Question

(A) The elevation in boiling point is given by the formula $\Delta T_{b} = i 	imes K_{b} 	imes m$. Since $K_{b}$ and $m$ are constant for the given

Vedclass · Accepted Answer

$CH_{3}COOH < NaCl < Na_{2}SO_{4} < K_{3}PO_{4}$

Vedclass · Answer

(A) The elevation in boiling point 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,$\Delta T_{b} \propto i$ (van't Hoff factor).$CH_{3}COOH$ is a weak electrolyte and dissociates partially,so its $i$ value is approximately $1 $NaCl$ dissociates into $2$ ions $(Na^+, Cl^-)$,so $i \approx 2$.$Na_{2}SO_{4}$ dissociates into $3$ ions $(2Na^+, SO_{4}^{2-})$,so $i \approx 3$.$K_{3}PO_{4}$ dissociates into $4$ ions $(3K^+, PO_{4}^{3-})$,so $i \approx 4$.Therefore,the order of boiling points is $CH_{3}COOH < NaCl < Na_{2}SO_{4} < K_{3}PO_{4}$.

The boiling point of $0.001 \ M$ aqueous solutions of $NaCl$,$Na_{2}SO_{4}$,$K_{3}PO_{4}$ and $CH_{3}COOH$ should follow the order.

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