Formaldehyde polymerizes to form glucose according to the reaction, `6HCHO rarr C_(6)H_(12)O_(6)` The theoretically computed equilibrium constant for this reaction is found to be `6 xx 10^(22)` If 1M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be :

Explanation

` 6 HCHO hArr C_(6) H_(12)O_(6)`
As equilibrium constant for this reaction is very large, therfore for the reverse reaction involving dissociation of glucose, equilibrium constant is very very small . Hence , for the reverse reaction
` C_(6) H_(12)O_(6) hArr 6 HCHO " " (K= 1/(6 xx 10^(22)))^(1//6)`
dissociation of glucose is negligible . Starting with 1 M, concentration at equilibrium at equilibrium `cong 1 M`
` K= ([HCHO]^(6))/([C_(6)H_(12)O_(6) )`
` 1/(6 xx 10^(22)) = ([HCHO]^(6))/1 or [HCHO] = (1/(6xx10^(22)))^(1//6)`
`log [HCHO] = 1/6 [-log (6xx10^(22)] = 1/6 [-22* 778 ] = -3* 7963 = bar 4* 2137`
` :. [HCHO] = 1* 636 xx 10^(-4) "M"`