On treatment of 100 ml of 0.1 M solution of the complex CrCl 3 .6H 2 O with excess of AgNO 3 , 4.305 g of AgCl was obtained. The complex is

Mol of AgCl = 4 . 3 0 5 1 4 3 . 5 = 0 . 0 3 = mol of Cl - given by the complex. Mol of the complex = 100 x 10 -3 x 0.1 = 0.01 ; Cr H 2 O 6 Cl 3 ⟶ Cr H 2 O 6 3 + + 3 Cl - 0 . 0 1 mol 0 . 0 1 mol 0 . 0 3 mol

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ChemistryCoordination ChemistryMiscellaneous/MixedEasy2 minPYQ_2020

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On treatment of 100 ml of 0.1 M solution of the complex CrCl₃.6H₂O with excess of AgNO₃, 4.305 g of AgCl was obtained. The complex is

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Answer:

Solution:

Mol of AgCl $= \frac{4.305}{143.5} = 0.03$ = mol of Cl^-given by the complex.

Mol of the complex = 100 x 10^-3x 0.1 = 0.01 ;

$\begin{matrix} [Cr {(H_{2} O)}_{6}] {Cl}_{3} & ⟶ & {[Cr {(H_{2} O)}_{6}]}^{3 +} & + & 3 {Cl}^{-} \\ 0.01 mol & 0.01 mol & 0.03 mol \end{matrix}$

Stream:NTA_ABHYASSubject:ChemistryTopic:Coordination ChemistrySubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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