A complex is represented as ${CoCl}_{3} x {NH}_{3^{-}}$ lts 0.1 molal solution in water shows $- {ΔT}_{f} = 0 .558 K . K_{f}$ for $H_{2} O$ is 1.86 K molality $^{- 1}$ . Assuming 100% ionization of complex and co-ordination number of Co is six, calculate formula of complex.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

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

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

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

None of these

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

$- {ΔT}_{f} = K_{f} x molality x i; 0.558 = 1.86 x 0.1 x i or i = 3$ Thus complex is $[{CoClxNH}_{3}] {Cl}_{2}$

since co-ordination number of CO is six, thus complex is $[CO {({NH}_{3})}_{5} Cl] {Cl}_{2}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE