The limiting molar conductance ${\lambda }_m^0$  of HCl, NaCl and  $C{H}_{3}COONa$ is given by 430, 120 & 90 S $c{m}^{2}mo{l}^{-1}$  respectively. If conductance of a 0.1 M solution of  $C{H}_{3}COOH$ is  $4\times {10}^{-3}$ S in a conducting cell of cell constant $\frac{1}{10}c{m}^{-1}$  then identify the correct statement (S)

${\lambda }_m^0=430+90-120$

A)  $R=\frac{l}{A}K=h\times \frac{l}{A}=-4\times {10}^{-3}\times \frac{1}{10}$

$K=4\times {10}^{-4}$

(B)  ${\lambda }_m=\frac{1000\times K}{C}=\frac{1000\times 4\times {10}^{-4}}{0.1}=4$
 $\alpha \%=\frac{{\lambda }_m}{{\lambda }_m^0}\frac{4}{400}\times 100=1\%$
(C)  $\underset{c(1-\alpha )}{C{H}_3}COOH\left(aq\right)\rightleftharpoons \underset{c\alpha }{C{H}_3}CO{O}^{-}\left(aq\right)+\underset{c\alpha }{H^{+}}$
 $\left[H^{+}\right]=c\alpha ={10}^{-1}\times {10}^{-2}={10}^{-3}$
(D)  $k_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{c{\alpha }^2}{1}$
$k_a=0.1\times {10}^{-4}=c_2{\alpha }_2^2 [\therefore 1-a\cong 1]$

$$\begin{gathered} {\alpha }_2=0.707\times {10}^{-2} \\ {\lambda }_m={\lambda }_m^0=400\times 0.707\times {10}^{-2}=\frac{k\text{'}\times 1000}{0.2} \\ k\text{'}=5.6\times {10}^{-4}sc{m}^{-1} \end{gathered}$$