A) \[16.875\times {{10}^{-12}}\]
B) \[16.875\times {{10}^{-14}}\]
C) \[1.6875\times {{10}^{-12}}\]
D) \[168.75\times {{10}^{-12}}\]
Correct Answer: C
Solution :
\[A{{g}_{2}}Cr{{O}_{4}}\underset{1.5\times {{10}^{-4}}}{\mathop{2A{{g}^{+}}}}\,+\underset{0.75\times {{10}^{-4}}}{\mathop{CrO_{4}^{2-}}}\,\] \[{{K}_{SP}}=\left[ A{{g}^{+}} \right]\,\,\left[ CrO_{4}^{2-} \right]\] \[{{\left[ 1.5\times {{10}^{-4}} \right]}^{2}}\,\,\left[ 0.75\times {{10}^{-4}} \right]\] \[=1.6875\times {{10}^{-12}}\]
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