JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
\( \wedge _m^o\) for \(NaCl,\,HCl\) and \(NaA\) are \(126.4,\,425.9\) and \(100.5\,\,S\,cm^2\,mol^{-1},\) respectively. If the conductivity of \(0.001\,M\,HA\) is \(5\times 10^{-5}\,S\,cm^{-1},\) degree of dissociation of \(HA\) is
- A \(0.50\)
- B \(0.25\)
- C \(0.125\)
- D \(0.75\)
Answer & Solution
Correct Answer
(C) \(0.125\)
Step-by-step Solution
Detailed explanation
\(\lambda _{m}^{o}(HA)\,=\,100.5\,+\,425.9\,-\,126.4\,=400\) \(\lambda _{m}^{o}=\,\frac{K\,\times \,1000}{M}\,=\,\frac{5\,\times \,{{10}^{-5}}\,\times \,{{10}^{3}}}{{{10}^{-3}}}\,=50\) \(\alpha \,=\,\frac{50}{400}\,=\,0.125\)
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