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Concentration Terms

The freezing ...

Question

The freezing point depression of 0.001m Kx[Fe(CN)6] is 7.10?10^-3 K. Determine the valueof x. Kf =1.86K kg /mol for water.

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Solution

$F r e e z i n g p o i n t d e p r e s s i o n ∆ T_{f} = i \times K_{f} \times m i = \frac{∆ T_{f}}{K_{f} \times m} = \frac{7.10 \times 10^{- 3}}{1.86 \times 0.001} = 3.82 \approx 4 T h e t o t a l n u m b e r o f v a l u e o f i = 4 T h e g i v e n c o m p l e x K_{x} [F e (C N)_{6}] h a s a l r e a d y c o m p l e x i o n [F e (C N)_{6}] a s 1 . S o t h e r e m a i n i n g a r e K i o n s = 4 - 1 = 3 T h u s t h e v a l u e o f x = 3 a n d t h e c o m p l e x i s K_{3} [F e (C N)_{6}]$

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