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(B) The molar mass of $HBr = 1 + 80 = 81 \ g \ mol^{-1}$.Number of moles of $HBr = \frac{8.1 \ g}{81 \ g \ mol^{-1}} = 0.1 \ mol$.Molality $(m) = \fra

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$- 3.53$

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(B) The molar mass of $HBr = 1 + 80 = 81 \ g \ mol^{-1}$.Number of moles of $HBr = \frac{8.1 \ g}{81 \ g \ mol^{-1}} = 0.1 \ mol$.Molality $(m) = \frac{0.1 \ mol}{0.1 \ kg} = 1 \ m$.For $HBr ightleftharpoons H^{+} + Br^{-}$,the degree of dissociation $\alpha = 0.9$.Van't Hoff factor $(i) = 1 + \alpha = 1 + 0.9 = 1.9$.Depression in freezing point $\Delta T_f = i 	imes K_f 	imes m = 1.9 	imes 1.86 	imes 1 = 3.534 \ K$.Freezing point of solution $T_f = T_f^{\circ} - \Delta T_f = 0 - 3.534 = - 3.534 \ ^oC$.

What is the freezing point of a solution containing $8.1 \ g \ HBr$ in $100 \ g$ water,assuming the acid to be $90 \ \%$ ionised ($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$)?

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