an electrolyte AB is 50% ionised in aqueous solution. Calculate the freezing point of 1 molal aqueous solution

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Solution

$T h e e l e c t r o l y t e A B d i s s o c i a t e s t o A B \leftrightarrow A^{+} + B^{-} I n i t i a l n o o f 1 0 0 m o l e s A t e q m 1 - α α α i = 1 - α + α + α = 1 + α A s α i s 50 % i o n i s e d = 0.5 s o i = 1 + 0.5 = 1.5 ∆ T_{f} = i K_{f} m = 1.5 \times 1.86 \times 1 = 2.79^{0} C T_{f} - T_{{f_{(}}_{S)}} = 2.79 0 - 2.79 = T_{{f_{(}}_{S)}} T_{{f_{(}}_{S)}} = - 2.79^{0} C T h u s t h e f r e e z i n g p o i n t o f s o l u t i o n = - 2.79^{0} C$

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A 0.2 molal aqueous solution of a weak acid (HX) is 20% ionised. The freezing point of this solution is (Given for water)

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