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Q.
If degree of dissociation of pure water at $100^{\circ} C$ is $1.8 \times 10^{-8}$, then the dissociation constant of water will be (density of $H _2 O =1 \,g / cc$ )
Equilibrium
Solution:
As, molarity, $=\frac{\text { wt. of solute per litre of solution }}{\text { Mol. wt. of solute }}$
Molarity of $H _2 O =\frac{1000}{18} $ mole / litre
$\underset{c(1-\alpha)}{H _2 O} \rightleftharpoons \underset{c\alpha}{H ^{+}}+ \underset{c\alpha}{OH ^{-} }$
Thus, $K_a=\frac{c \alpha^2}{1-r}=c \alpha^2=1.8 \times 10^{-14}$