\(\Lambda_m^\circ\) for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 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 -
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- 0.50
- 0.125
- 0.25
- 0.75
The Correct Option is B
Solution and Explanation
= 425.9 + 100.5 − 126.4 = 400 S cm2 mol−1.
Molar conductivity of 0.001 M HA: \(\Lambda_m=\dfrac{\kappa\times1000}{C}=\dfrac{5\times10^{-5}\times1000}{0.001}=50\) S cm2 mol−1.
Degree of dissociation: \(\alpha=\dfrac{\Lambda_m}{\Lambda_m^\circ}=\dfrac{50}{400}=0.125\).
Answer: (2) 0.125
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