A) \[4{{s}^{3}}\]
B) \[{{s}^{3}}\]
C) \[2{{s}^{3}}\]
D) \[16{{s}^{2}}\]
Correct Answer: A
Solution :
\[A{{g}_{2}}Cr{{O}_{4}}\]ionises completely in the solution as \[\underset{s\,\,mol\text{/}L}{\mathop{A{{g}_{2}}Cr{{O}_{4}}}}\,\xrightarrow{{}}\underset{2s\,\,mol\text{/}L}{\mathop{2A{{g}^{+}}}}\,+\underset{s\,\,mol\text{/}L}{\mathop{CrO_{4}^{2-}}}\,\] Hence, the solubility product, \[{{K}_{sp}}\]of\[A{{g}_{2}}Cr{{O}_{4}}={{[A{{g}^{+}}]}^{2}}[CrO_{4}^{2-}]\] \[={{(2s)}^{2}}(s)\] \[=4{{s}^{3}}\]
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