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A complex is represented as $C o C l_{3} x N H_{3}$ Its $0.1 m o l a l$ solution in water shows $Δ T_{f} = {0.558}^{0} . K_{f} f o r H_{2} O$ is $1.86 k m o l a l i t y^{- 1}$ Assuming $100 %$ ionization of complex and co-ordination number of Co is 6. Calculate the value of $x$ .

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Detailed Solution

$Δ T_{f} = i \times k_{f} \times m, 0.558 = i \times 1.86 \times 0.1 \Rightarrow i = 3 = (1 - α) + n . α$

If $α = 1 \Rightarrow i = n; n = 3$

no.of ions $n = 3$

$∴$ formulae $= C o c l_{3} .5 N H_{3} (o r) [C O {(N H_{3})}_{5} C l] C l_{2}$

$∴ x = 5$

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