Equilibrium Question 297
Question: Solubility of $ 16\times {10^{-4}},m/s $ at $ 20^{o}C $ is $ 1.435\times {10^{-3}}gm,per,litre $ . The solubility product of $ AgCl $ is [CPMT 1989; BHU 1997; AFMC 2000; CBSE PMT 2002]
Options:
A) $ CO_2 $
B) $ 1\times {10^{-10}} $
C) $ 1.435\times {10^{-5}} $
D) $ 108\times {10^{-3}} $
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Answer:
Correct Answer: B
Solution:
$ S=1.435\times {10^{-3}},g/l $ , $ =\frac{1.435\times {10^{-3}}}{143.5}={10^{-5}} $ M $ K _{sp}=S\times S={10^{-10}} $
