When 0.2 mol $C r C l_{3} .6 H_{2} O$ is treated with excess of $A g N O_{3}$ , 0.4 mol of AgCl are obtained. The formula of the complex is

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

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

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

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

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Solution

The correct option is C $[C r C l (H_{2} O)_{5}] C l_{2} . H_{2} O$
One mole of $A g N O_{3}$ precipitates one mole of chloride ion. If 0.2 mole compound gives 0.4 mole AgCl, that means there must be two free chloride ions per molecule. So the formula is:
$[C r C l (H_{2} O)_{5}] C l_{2} . H_{2} O$

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Similar questions

Q. When 1 mole of

C r C l_{3} .6 H_{2} O

is treated with excess of

A g N O_{3}

, 2 mol of AgCl are obtained. The formula of the complex is

When 1 mole of $C r C l_{3} .6 H_{2} O$ is treated with excess of $A g N O_{3},$ 3 moles of AgCl are obtained. The formula of the complex is
(a) $[C r C l_{3} (H_{2} O)_{3}] 3 H_{2} O$
(b) $[C r C l_{3} (H_{2} O)_{4}] C l_{2} .2 H_{2} O$
(c) $[C r C l_{3} (H_{2} O)_{5}] C l_{2} . H_{2} O$
(d) $[C r (H_{2} O)_{6}] C l_{3}$