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Raoult's Law

An aqueous so...

Question

An aqueous solution containing 12.48g of barium chloride in 1.0kg of water boils at 373.0832K. Calculate the degree of dissociation of barium chloride. (Given Kb for water= 0.52K/m)

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Solution

$A c c o r d i n g t o e l e v a t i o n o f b o i l i n g p o i n t, ∆ T_{b} = K_{b} \times m = K_{b} \times \frac{W_{2}}{M_{2} \times W_{1}} ∆ T_{b} - e l e v a t i o n o f b o i l i n g p o i n t K_{b} - m o l a l e l e v a t i o n c o n s \tan t W_{2} - W e i g h t o f t h e s o l u t e M_{2} - M o l e c u l a r m a s s o f t h e s o l u t e W_{1} - W e i g h t o f t h e s o l v e n t ∆ T_{b} = K_{b} \times \frac{W_{2}}{M_{2} \times W_{1}} \times 1000 = 0.52 \times \frac{12.48}{208.23 \times 1} = 0.0312 ∆ T_{b} (g i v e n) = T_{b} - {T_{b}}^{0} = 373.0832 - 373 = 0.0832 D e g r e e o f d i s s o c i a t i o n α = \frac{i - 1}{n - 1} W h e r e i = V a n t H o f f f a c t o r i = \frac{O b s e r v e d c o l l i g a t i v e p r o p e r t y}{C a l c u l a t e d c o l l i g a t i v e p r o p e r t y} = \frac{0.0832}{0.0312} = 2.67 α = \frac{2.67 - 1}{3 - 1} (w h e r e n = 3 a s B a C l_{2} d i s s o c i a t e s B a C l_{2} \to B a^{2 +} + 2 C l^{-} = \frac{1.67}{2} = 0.8333 = 0.8333 \times 100 = 83.33 %$

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