Question

Here is the question:-

☆☆Degree of dissociation of 0.1 N $\bold{CH_3COOH}$ is 【if dissociation constant =1×10^(-5)】

Answer 1

Concentration :- 0.1N

$normality = molarity \times n - factor \\ 0.1 = molarity \times 1 \\ molarity = 0.1$
Molarity=0.1 M

we have to find degree of dissociation
$\alpha$
since degree of dissociation is ≤ 10^(-3)

so,

$k = \frac{ { \alpha }^{2}c }{1 - \alpha } \\ but \: k \leqslant {10}^{ - 3} \\ hence \: \alpha \: will \: be \: taken \: as \: negligible \\ {10}^{ - 5} = \frac{ { \alpha }^{2} \times 0.1 }{1} \\ {10}^{ - 4} = { \alpha }^{2} \\ \alpha = {10}^{ - 2}$

Hope it will help you‼️‼️

Answer 2

Heyaa..

ur ans is in this attachment..

hope it helps..!!