A) \[{{10}^{-4}}\]
B) \[0.01\]
C) \[0.001\]
D) \[0.1\]
Correct Answer: A
Solution :
\[\underset{s}{\mathop{AgCl}}\,\underset{s}{\mathop{A{{g}^{+}}}}\,+\underset{s}{\mathop{C{{l}^{-}}}}\,\] Solubility \[=1.435\text{ }g/litre\] Solubility \[\text{=}\frac{1.435}{143.5}\text{ }mole/litre\] \[143.5=\]molecular weight Solubility product \[={{(s)}^{2}}\] \[={{(0.01)}^{2}}={{10}^{-4}}\]
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