JEE PYQ: Chemical Equilibrium Question 28
Question 28 - 2019 (12 Jan Shift 1)
$\Lambda_m^\circ$ 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:
(1) 0.50 (2) 0.25 (3) 0.125 (4) 0.75
Show Answer
Answer: (3) 0.125
Solution
$\Lambda_m^\circ(\text{HA}) = 425.9 + 100.5 - 126.4 = 400$. $\Lambda_m = 50$. $\alpha = 50/400 = 0.125$.
BETA
