A) \[4.32\times {{10}^{-18}}\]
B) \[1.8\times {{10}^{-9}}\]
C) \[8.64\times {{10}^{-13}}\]
D) None of these
Correct Answer: A
Solution :
\[\lambda =K\times \frac{1000}{N}\] \[1.50\times {{10}^{-4}}\times {{10}^{4}}=9\times {{10}^{-6}}\times {{10}^{-2}}\times \frac{1000}{N}\] \[N=6\times {{10}^{-5}}\] \[S=M=\frac{N}{{{n}_{f}}}=\frac{6\times {{10}^{-5}}}{3}=2\times {{10}^{-5}}mol/L\] \[Ag\underset{3S}{\mathop{_{3}P{{O}_{4}}}}\,=3\underset{S}{\mathop{A{{g}^{+}}+}}\,P{{O}_{4}}\] \[{{K}_{sp}}={{(3S)}^{3}}.S=27{{S}^{4}}=27\times {{(2\times {{10}^{-5}})}^{4}}=4.32\times {{10}^{-18}}\]
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