The degree of dissociation of a 0.2 , m aqueo | vedclass

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Question

(D) For the dissociation of weak acid $HX$: $HX \rightleftharpoons H^+_{(aq)} + X^-_{(aq)}$Initial moles: $1, 0, 0$Equilibrium moles: $(1 - 0.3), 0.3,

Vedclass · Accepted Answer

$-0.48$

Vedclass · Answer

(D) For the dissociation of weak acid $HX$: $HX ightleftharpoons H^+_{(aq)} + X^-_{(aq)}$Initial moles: $1, 0, 0$Equilibrium moles: $(1 - 0.3), 0.3, 0.3$Total moles at equilibrium $= 1 - 0.3 + 0.3 + 0.3 = 1.3$Van't Hoff factor $i = \frac{	ext{Total moles at equilibrium}}{	ext{Initial moles}} = \frac{1.3}{1} = 1.3$Depression in freezing point $\Delta T_f = i 	imes K_f 	imes m$$\Delta T_f = 1.3 	imes 1.85 	imes 0.2 = 0.481 \, ^oC$Freezing point of solution $= T_f^o - \Delta T_f = 0 \, ^oC - 0.481 \, ^oC = -0.481 \, ^oC \approx -0.48 \, ^oC$.

The degree of dissociation of a $0.2 \, m$ aqueous solution of a weak acid $HX$ is $0.3$. If $K_f$ for water is $1.85 \, K \, kg \, mol^{-1}$,then the freezing point of the solution will be approximately ........... $^oC$.

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