A) \[3.3\times {{10}^{2}}mol\,{{L}^{-1}}\]
B) \[3\times {{10}^{-1}}mol\,{{L}^{-1}}\]
C) \[3\times {{10}^{-3}}mol\,{{L}^{-1}}\]
D) \[3\times {{10}^{3}}mol\,{{L}^{-1}}\]
Correct Answer: C
Solution :
\[{{K}_{c}}=\frac{{{\left[ N{{O}_{2}} \right]}^{2}}}{\left[ {{N}_{2}}{{O}_{4}} \right]}=\frac{{{(1.2\times {{10}^{-2}})}^{2}}}{4.8\times {{10}^{-2}}}\] \[=3\times {{10}^{-3}}mol\,\,{{L}^{-1}}\]You need to login to perform this action.
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