Question

ꓥm of 0.05 M weak electrolyte is 50 Sm2mol-1, ꓥm0of it is 440 Sm2mol-1. Calculate α (degree of dissociation) of the electrolyte.​

Answer 1

Answer:

0.114

Explanation:

$\alpha = \frac{ \lambda {}^{0}m }{ \lambda {}^{c} m}$

$\alpha = \frac{50}{440} \\ \alpha = \frac{5}{44} \\ \alpha = 0.114(approx)$