In a solvent,50,% of an acid HA dimerizes and | vedclass

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(A) Let the initial moles of $HA$ be $a = 1 	ext{ mole}$.$50\,\%$ of $HA$ $(0.5 	ext{ mole})$ dimerizes: $2HA ightleftharpoons H_2A_2$.Final moles

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(A) Let the initial moles of $HA$ be $a = 1 	ext{ mole}$.$50\,\%$ of $HA$ $(0.5 	ext{ mole})$ dimerizes: $2HA ightleftharpoons H_2A_2$.Final moles for dimerization: $0.5 - 0.25 = 0.25 	ext{ mole of } HA$ and $0.25 	ext{ mole of } H_2A_2$.$50\,\%$ of $HA$ $(0.5 	ext{ mole})$ dissociates: $HA ightleftharpoons H^+ + A^-$.Final moles for dissociation: $0.5 	ext{ mole of } H^+$ and $0.5 	ext{ mole of } A^-$.Total final moles $= 0.25 (HA) + 0.25 (H_2A_2) + 0.5 (H^+) + 0.5 (A^-) = 1.5 	ext{ moles}$.Van't Hoff factor $i = \frac{	ext{Total final moles}}{	ext{Initial moles}} = \frac{1.5}{1} = 1.5$.$1.5 = 150 	imes 10^{-2}$.

In a solvent,$50\,\%$ of an acid $HA$ dimerizes and the rest dissociates. The van't Hoff factor of the acid is $.....\times 10^{-2}$. (Round off to the nearest integer)

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