Question

conductivity of 0.00241M acetic acid solution is 7.896×10^5S cm^-1 . Calculate it's molar conductivity in this solution.​

Answer 1

Explanation:

ANSWER

Given: K=7.896×10

−5

Scm

−1

C=M=0.00241molL

−1

∴ Molar conductivity

∧m=(K×1000)/M=(7.896×10

−5

×1000)/0.00241

=32.76Scm

2

mol

−1

Degree of dissociation

α=

λ

o

m

λm

=

390.5

32.76

=0.084

∴ Dissociation constant

⇒K=

1−α

Cα

=

(1−0.084)

0.00241×0.084

=1.86×10

−5

molL

−1