A six coordinate complex of formula ${CrCl}_{3} . 6 H_{2} O$ has green colour. One liter of 0.1 M solution of the complex when treated with excess of $A g N O_{3}$ gave 28.7 g of white precipitate. The formula of the complex would be :

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$[C r C l_{2} {(H_{2} O)}_{4}] C l .2 H_{2} O$

$[C r {(H_{2} O)}_{3} C l_{3}]$

$[C r C l {(H_{2} O)}_{5}] C l_{2} H_{2} O$

$[C r {(H_{2} O)}_{6}] C l_{3}$

answer is B.

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Detailed Solution

No. of moles of AgCl formula $= \frac{28.7}{143.5} = 0.2$

$∴ 0.1$ mole complex then 0.2 moles of free $C l^{-}$

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