JEE Main 2019 Chemistry question
for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm mol, respectively. If the conductivity of 0.001 M HA is S cm, degree of dissociation of HA is:
Answer: (B)
Solution
(HA) = 425.9 + 100.5 − 126.4 = 400; = 5 × 10 × 10 = 50; α = 50/400.