Question

Treatment of 100 ml of 0.1 m solution of cocl3.6h2o with excess agno3; 1.2 Ă— 1022 ions are precipitated. The complex is

Answer 1

Treatment of 100 ml of 0.1 m solution of cocl3.6h2o with excess agno3; 1.2 Ă— 1022 ions are precipitated. The complex is
=
1.2 Ă— 1022

100 ml of 0.1 m solution of cocl3.6h2o
=0.9

Answer 2

Hi there dear user☺️

=> On treatment of 100 ml of 0.1 M solution of CoCl3 . 6H2O with access AgNO3; 1.2 × 10²² ions are participated. The complex is

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⏬ANSWER⏬

[Co(H2O)5Cl]Cl2.H2O

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⤵️SOLUTION⤵️

Moles of complex = Molarity × volume (ml) / 1000

= 100 × 0.1 / 1000

= 0.01 mole

Moles of ions precipitated with excess of

AgNO3 = 1.2 × 10²² / 6.02 × 10²³ = 0.02 moles

Number of Cl present in ionization sphere =

Mole of ion precipitated with excess AgNO3 / Mole of complex = 0.02 / 0.01

= 2

It means 2Cl ions present in ionization sphere

Therefore, complex is [Co(H2O)5Cl]Cl2.H2O

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(check the attachment as well if you still have doubt)

Hope it helps you out✌️

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