A solution containing $2 .675 g of {CoCl}_{3} \cdot 6 {NH}_{3}$ (molar mass = 267.5 g mol-1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of ${AgNO}_{3} to give 4 .78 g of AgCl (molar mass = 143 .5 g$ ${mol}^{- 1})$ The formula of the complex is (At. mass of Ag = 108 u)

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$[CoCl {({NH}_{3})}_{5}] {Cl}_{2}$

$[{CoCl}_{3} {({NH}_{3})}_{3}]$

$[{CoCl}_{2} {({NH}_{3})}_{4}] Cl$

$[Co {({NH}_{3})}_{6}] {Cl}_{3}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\underset{2 .675 g}{{CoCl}_{3} . 6 {NH}_{3}} \to \times {Cl}^{-} {xCl}^{-} + {AgNO}_{3} ⟶ \underset{4 .78 g}{AgCl}$

Number of moles of the complex
$= \frac{2 .675}{267 .5} = 0 .01 moles$
Number of moles of AgCl obtained
$= \frac{4 .78}{143 .5} = 0 .03 moles$

$∴$ No. of moles of AgCl obtained by 1mole of complex is,
$∴ n = \frac{0 .03}{0 .01} = 3$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

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

Take Free Test