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(B) The dissociation reaction is: $HBr \rightleftharpoons H^{+} + Br^{-}$Initial moles: $1, 0, 0$Moles at equilibrium: $(1-\alpha), \alpha, \alpha$Giv

Vedclass · Accepted Answer

$-3.53$

Vedclass · Answer

(B) The dissociation reaction is: $HBr ightleftharpoons H^{+} + Br^{-}$Initial moles: $1, 0, 0$Moles at equilibrium: $(1-\alpha), \alpha, \alpha$Given degree of ionization $\alpha = 0.9$.Total moles at equilibrium $= 1 - \alpha + \alpha + \alpha = 1 + \alpha = 1 + 0.9 = 1.9$.Van't Hoff factor $i = 1.9$.Molality $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{8.1 / 81}{100 / 1000} = \frac{0.1}{0.1} = 1 \ m$.Depression in freezing point $\Delta T_f = i 	imes K_f 	imes m = 1.9 	imes 1.86 	imes 1 = 3.534 \ ^\circ C$.Freezing point of solution $= 0 - \Delta T_f = 0 - 3.534 \approx -3.53 \ ^\circ C$.

The freezing point of a solution prepared by adding $8.1 \ g$ of $HBr$ ($90\%$ ionized) to $100 \ g$ of water is $...... \ ^\circ C$. (Given $K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$)

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