Formaldehyde polymerizes to form glucose according to the reaction 6HCHOiffC_(6)H_(12)O_(6) The theoretically computed equilibrium constant for this reaction is found to be 6xx10^(22). If 1 M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be

Formaldehyde polymerizes to form glucose according to the reaction 6H

1.	Formaldehyde polymerizes to form glucose according to the reaction 6HCHOiffC_(6)H_(12)O_(6) The theoretically computed equilibrium constant for this reaction is found to be 6xx10^(22). If 1 M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be
Answer» `1.6XX10^(-2)M` `1.6xx10^(-4)M` `1.6xx10^(-6)M` `1.6xx10^(-8)M` Solution :A very high vallue 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)]=1M` `K=([C_(6)H_(12)O_(6)])/([HCHO]^(6))i.e.,6xx10^(22)=(1)/([HCHO]^(6))` or `[HCHO]=((1)/(6xx10^(22)))^(1//6)=1.6xx10^(-4)M`