The freezing point of a 0.2 m aqueous soluti | vedclass

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(B) For a weak acid $HX$,the dissociation is $HX \rightleftharpoons H^+ + X^-$. Initial moles: $1, 0, 0$. Moles at equilibrium: $(1-\alpha), \alpha, \

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$-0.45$

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(B) For a weak acid $HX$,the dissociation is $HX ightleftharpoons H^+ + X^-$. Initial moles: $1, 0, 0$. Moles at equilibrium: $(1-\alpha), \alpha, \alpha$. Total moles at equilibrium = $1-\alpha + \alpha + \alpha = 1+\alpha$. Van't Hoff factor $i = 1+\alpha$. Given $\alpha = 20\% = 0.2$. So,$i = 1 + 0.2 = 1.2$. The depression in freezing point is given by $\Delta T_f = i 	imes K_f 	imes m$. $\Delta T_f = 1.2 	imes 1.86 	imes 0.2 = 0.4464 \ ^oC$. Since $\Delta T_f = T_f^0 - T_f$,and $T_f^0 = 0 \ ^oC$ for water,$0.4464 = 0 - T_f \implies T_f = -0.4464 \ ^oC \approx -0.45 \ ^oC$.

The freezing point of a $0.2 \ m$ aqueous solution of a weak acid $(HX)$ which is $20\%$ ionized is ......... $^oC$. (Given: $K_f = 1.86 \ ^oC/m$ for water)

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