Question

A $1 M$ solution of glucose reaches dissociation equilibrium given below.
$C _{6} H _{12} O _{6} \rightleftharpoons 6 HCHO$
If the equilibrium constant is $0.167 \times 10^{-22}$, the concentration of $HCHO$ in the equilibrium is

• A. $1.60 \times 10^{-8} M$
• B. $3.20 \times 10^{-6} M$
• C. $3.20 \times 10^{-4} M$
• D. $1.60 \times 10^{-4} M$

Answer 1

Answer: (D)

$K =\frac{[ HCHO ]^{6}}{\left[ C _{6} H _{12} O _{6}\right]}=\frac{(6 x )^{6}}{(1- x )} \simeq \frac{(6 x )^{6}}{1}$
$0.167 \times 10^{-22}=(6 x )^{6}$
$6 x =\left(0.167 \times 10^{-22}\right)^{1 / 6}=\left[0.167 \times 10^{2} \times 10^{-24}\right]^{1 / 6}$
$=[16.7]^{1 / 6} \times 10^{-4}$
$=1.6 \times 10^{-4}$
$\left[\right.$ because $\left.[1.6]^{6}=[2.56]^{3}=6.4 \times 2.56=16.8\right]$