Question

Formaldehyde polymerizes to form glucose according to the reaction $6 HCHOC _{6} H _{12} O _{6}$ The theoretically computed equilibrium constant for this reaction is found to be $6 \times 10^{22}$ . If $1\, M$ solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be

• 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 $

Answer 1

Answer: (B)

A very high value of $K$ for the given equilibrium shows that dissociation of glucose to form $H C H O$ is very-very small.
Hence, at equilibrium, we can take,
$\left[ C _{6} H _{12} O _{6}\right]=1 M $
$K =\frac{\left[ C _{6} H _{12} O _{6}\right]}{[ HCHO ]^{6}} $
i.e, $6 \times 10^{22}=\frac{1}{[ HCHO ]^{6}} $
or $[ HCHO ]=\left(\frac{1}{6 \times 10^{22}}\right)^{1 / 6}$
$=1.6 \times 10^{-4} M$