JEE Main Solved Paper 2017 Question 25
Question: On treatment of 100 mL of 0.1 M solution of $ CoCl _3.6H _2O $ with excess $ AgNO _3;1.2\times 10^{22} $ ions are precipitated. The complex is :- [JEE Main Solved Paper-2017]
Options:
A) $ AgNO _3;1.2\times 10^{22} $
B) $ [Co{{(H _2O)}_3}Cl _3].3H _2O $
C) $ [Co{{(H _2O)}_6}]Cl _3 $
D) $ [Co{{(H _2O)}_5}Cl]Cl _2.H _2O $
Show Answer
Answer:
Correct Answer: D
Solution:
$ Moles,of,complex,\text{=},\frac{Molarity\times volume,\text{(ml)}}{1000} $ $ =\frac{100\times 0.1}{1000}=0.01,mole $
Moles of ions precipitated with excess of $ AgNO _3=\frac{1.2\times 10^{22}}{6.02\times 10^{23}} $ $ =0.02,moles $
Number of $ C{l^{-}} $ present in ionization sphere = $ \frac{Mole,of,ion,precipitated,with,exess,AgN{O_3}}{Mole,of,complex}=\frac{0.02}{0.01}=2 $
It means $ 2C{l^{-}} $ ions present in ionization sphere
$ \therefore $ complex is $ [\text{ Co(}{H_2}O{{)}_5}\text{Cl }]\text{ C}{l_2}\text{.}{H_2}O $
