A) \[1.0\times {{10}^{-4}}\]g
B) \[28.3\times {{10}^{-2}}g\]
C) \[2.83\times {{10}^{-3}}g\]
D) \[1.0\times {{10}^{-7}}g.\]
Correct Answer: C
Solution :
[c] Let s = solubility \[AgI{{O}_{3}}\rightleftharpoons \underset{s}{\mathop{A{{g}^{+}}}}\,+\underset{s}{\mathop{IO_{3}^{-}}}\,\] \[{{K}_{sp}}=[A{{g}^{+}}][I{{O}_{3}}^{-}]=s\times s={{s}^{2}}\] Given \[{{K}_{sp}}=1\times {{10}^{-8}}\] \[\therefore s=\sqrt{{{K}_{sp}}}=\sqrt{1\times {{10}^{-8}}}\] \[=1.0\times {{10}^{-4}}mol/lit=1.0\times {{10}^{-4}}\times 283\text{ }g/lit\] (\[\therefore \]Molecular mass of\[AgI{{O}_{3}}=283\]) \[=\frac{1.0\times {{10}^{-4}}\times 283\times 100}{1000}g/100mL\] \[=2.83\times {{10}^{-3}}g/100mL\]
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