A solution contains 2.675g of $C o C l_{3} .6 N H_{3}$ (molar mass $= 267.5 g m o l^{- 1}$ ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $A g N O_{3}$ to give 4.78g of $A g C l$ (molar mass $= 143.5 g m o l^{- 1}$ ). The formula of the complex is (At.mass of $A g = 108 u$ )

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$[C o C l {(N H_{3})}_{5}] C l_{2}$

$[C o {(N H_{3})}_{6}] C l_{3}$

$[C o C l_{2} {(N H_{3})}_{4}] C l$

$[C o C l_{3} {(N H_{3})}_{3}]$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

No.of moles of $C o C l_{3} .6 N H_{3} = \frac{2.675}{267.5} = 0.01$
No.of moles of $A g C l = \frac{4.78}{143.5} = 0.03$

Since 0.01 moles of the complex $C o C l_{3} .6 N H_{3}$ gives 0.03 moles of $A g C l$ on treatment with $A g N O_{3}$ , it implies that 3 chloride ions are ionisable, in the complex. Thus, the formula of the complex is $[C o {(N H_{3})}_{6}] C l_{3}$ .