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

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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

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

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

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

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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