# formaldehyde polymerizes to form glucose according to the reaction 6HCHO<=>C6H12O6 . the theoretically calculated equilibrium constant for this reaction is found to be 6 x 10(to the power 22). if 1M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be?

Dear Student,

Please find below the solution to the asked query:

The dissociation of glucose to form formaldehyde at very high value of equilibrium constant is very small.

Therefore at equilibrium,

${\text{REACTION\hspace{0.17em}:\hspace{0.17em}6HCHO\hspace{0.17em}\u21d4C}}_{6}{H}_{12}{O}_{6}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[Glucose\right]=1M\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}K=\frac{\left[{C}_{6}{H}_{12}{O}_{6}\right]}{{\left[HCHO\right]}^{6}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}6\times {10}^{22}=\frac{1}{{\left[HCHO\right]}^{6}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[HCHO\right]={\left[\frac{1}{6\times {10}^{22}}\right]}^{\frac{1}{6}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[HCHO\right]=1.6\times {10}^{-4}M\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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