On treatment of 10 ml of 1 M solution of the complex $C r C l_{3} .6 H_{2} O$ with excess of $A g N O_{3}, 4.305$ g of AgCl was obtained. 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 D $[C r {(H_{2} O)}_{6}] C l_{3}$
No. of moles of given complex $= 10 \times 1 = 10$ millimole.
And no . of millimoles of AgCl formed $= (4.305 / 143.5) = 0.03$ mole $= 30$ millimole.
As no.of moles of AgCl formed $=$ no. of moles of chloride ion given by complex.
So, no. of moles of chloride ion given by complex $= 30$ millimole.
That means every mole of complex gives three moles of chloride ion.
So compound is - $[C r (H_{2} O)_{6}] C l_{3}$ .

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Q. When 1 mole of

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

is treated with excess of

A g N O_{3}

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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

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) $[C r C l_{3} (H_{2} O)_{4}] C l_{2} .2 H_{2} O$
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Q. Which isomer of

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

is dark green in colour and forms one mole of AgCl with excess of

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solution -

On treatment of 100 mL of 0.1M solution of $C o C l_{3} .6 H_{2} O$ with excess $A g N O_{3}; 1.2 \times 10^{22}$ ions are precipitated. The complex is

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On treatment of 10 ml of 1 M solution of the complex CrCl3.6H2O with excess of AgNO3,4.305 g of AgCl was obtained. The complex is :

On treatment of 10 ml of 1 M solution of the complex $C r C l_{3} .6 H_{2} O$ with excess of $A g N O_{3}, 4.305$ g of AgCl was obtained. The complex is :