Compound $P d C l_{4} .6 H_{2} O$ is a hydrated complex. One molal aqueous solution of it has freezing point 269.28 K. Assuming 100% ionisation of the complex, which of the following is the formula of the complex $(K_{f} f o r w a t e r = 1.86 K K g m o l^{- 1})$ ?

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$[P d {(H_{2} O)}_{3} C l_{3}] C l .3 H_{2} O$

$[P d {(H_{2} O)}_{4} C l_{2}] C l_{2} .2 H_{2} O$

$[P d {(H_{2} O)}_{6}] C l_{4}$

$[P d {(H_{2} O)}_{2} C l_{4}] .4 H_{2} O$

answer is C.

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Detailed Solution

$Δ T = i \times K f \times m (273 - 269.28) = i \times 1.86 \times 1 i = 2 α = \frac{i - 1}{n - 1} 1 = \frac{2 - 1}{n - 1} n = 2$
Thus, the complex is $[P d {(H_{2} O)}_{3} C l_{3}] C l .3 H_{2} O$

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