A) \[5.7\times {{10}^{-12}}\]
B) \[1.32\times {{10}^{-12}}\]
C) \[7.5\times {{10}^{-12}}\]
D) \[1.74\times {{10}^{-12}}\]
Correct Answer: D
Solution :
: Using the relationship between solubility of sparingly soluble salt and \[\wedge _{m}^{o}\], \[k=1.85\times {{10}^{-5}}S\,{{m}^{-1}}\,=1.85\times {{10}^{-7}}S\,c{{m}^{-1}}\] \[\wedge _{m}^{o}=140\times {{10}^{-4}}\,S\,{{m}^{2}}\,mo{{l}^{-1}}=140\,S\,c{{m}^{2}}\,mo{{l}^{-1}}\]\[s=\frac{k\times 1000}{\wedge _{m}^{o}}=\frac{1.85\times {{10}^{-7}}\times 1000}{140}\] \[s=\frac{k\times 1000}{\wedge _{m}^{o}}=\frac{1.85\times {{10}^{-7}}\times 1000}{140}\] \[=1.32\times {{10}^{-6}}\] \[{{K}_{sp}}={{(s)}^{2}}\] [For AB type of salt] \[={{(1.32\times {{10}^{-6}})}^{2}}=1.74\times {{10}^{-12}}\]
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