A) \[1.6\times {{10}^{-2}}M\]
B) \[1.6\times {{10}^{-4}}M\]
C) \[1.6\,\times {{10}^{-6}}M\]
D) \[1.6\,\times {{10}^{-8}}\,M\]
Correct Answer: B
Solution :
A very high value of K for the given equilibrium shows that dissociation of glucose to form HCHO is very-very small. Hence, at equilibrium, we can take, \[[{{C}_{6}}{{H}_{12}}{{O}_{6}}]=1\,M\] \[K=\frac{[{{C}_{6}}{{H}_{12}}{{O}_{6}}]}{{{[HCHO]}^{6}}}ie,6\times {{10}^{22}}=\frac{1}{{{[HCHO]}^{6}}}\] or \[[HCHO]={{\left( \frac{1}{6\times {{10}^{22}}} \right)}^{1/6}}=1.6\times {{10}^{-4}}M\]You need to login to perform this action.
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