JEE MAIN - Chemistry (2017 (Offline) - No. 11)
On treatment of 100 mL of 0.1 M solution of CoCl3. 6H2O with excess AgNO3; 1.2 $$\times$$ 1022 ions are precipitated. The complex is :
[Co(H2O)3Cl3].3H2O
[Co(H2O)6]Cl3
[Co(H2O)5Cl]Cl2.H2O
[Co(H2O)4Cl2]Cl.2H2O
Explanation
Moles of complex = $${{Molarity \times Volume\left( {mL} \right)} \over {1000}}$$
= $${{100 \times 0.1} \over {1000}}$$ = 0.01 mole
Moles of ions precipitated with excess of AgNO3
= $${{1.2 \times {{10}^{22}}} \over {6.023 \times {{10}^{23}}}}$$ = 0.02 mole
So 0.01 Ć n = 0.02
$$ \Rightarrow $$ n = 2
It means 2Clā ions present in ionization sphere.
$$ \therefore $$ The formula of complex is [Co(H2O)5Cl]Cl2.H2O
= $${{100 \times 0.1} \over {1000}}$$ = 0.01 mole
Moles of ions precipitated with excess of AgNO3
= $${{1.2 \times {{10}^{22}}} \over {6.023 \times {{10}^{23}}}}$$ = 0.02 mole
So 0.01 Ć n = 0.02
$$ \Rightarrow $$ n = 2
It means 2Clā ions present in ionization sphere.
$$ \therefore $$ The formula of complex is [Co(H2O)5Cl]Cl2.H2O
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