• # question_answer $\wedge _{m}^{o}$ for $NaCl,\text{ }HCl$ and $NaA$ are 126.4, 425.9 and 100.5 $S\,c{{m}^{2}}mo{{l}^{-1}},$ respectively. If the conductive of $0.001\text{ }MHA$ is $5\times {{10}^{-5}}S\,c{{m}^{-1}},$ degree of dissociation of HA is: A) 0.50 B) 0.25 C) 0.125                D) 0.75

 $\lambda _{m}^{0}(HA)=100.5+425.9-126.4=400$ $\lambda _{m}^{0}=\frac{K\times 100}{M}=\frac{5\times {{10}^{-5}}\times {{10}^{3}}}{{{10}^{-3}}}=50$ $\alpha =\frac{50}{400}=0.125.$