A) \[{{(2.22\times {{10}^{-31}})}^{1/4}}\]
B) \[{{(18\times {{10}^{-31}})}^{1/2}}\]
C) \[{{(18\times {{10}^{-31}})}^{1/4}}\]
D) \[{{(4.86\times {{10}^{-29}})}^{1/4}}\]
Correct Answer: C
Solution :
\[Cr{{(OH)}_{3}}\underset{S}{\mathop{C{{r}^{3+}}}}\,+{{\underset{3S}{\mathop{3O{{H}^{-}}}}\,}_{\text{ }\!\![\!\!\text{ S}\,\,\text{is}\,\,\text{solubility }\!\!]\!\!\text{ }}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,27{{S}^{4}}=6\times {{10}^{-31}}\] \[S={{\left( \frac{6}{27}\times {{10}^{-31}} \right)}^{1/4}}\] \[\because \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[O{{H}^{-}}]=3S\] \[=3{{\left( \frac{6}{27}\times {{10}^{-31}} \right)}^{1/4}}\] \[={{(18\times {{10}^{-31}})}^{1/4}}\]
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