Question

2
0
6. The conductivity of 0.025 mol L- methanoic acid is
46.1 s cm2 mol-!. Calculate the degree of
dissociation and dissociation constant. (Given, a
349.6 S cmmol-
= 54.6 S cm mol-
(HCOO)
0
20​

Answer 1

Explanation:

Answer:Given that

λ0(H+)= 349.6 S cm2 mol–1

λ0(HCOO–) = 54.6 S cm2 mol–1

Concentration ,C = 0.025 mol L-1

λ(HCOOH) = 46.1 S cm2 mol−1

use formula

λ°(HCOOH) = λ0(H+) + λ0(HCOO–)

plug the values we get

λ°(HCOOH) = 0.349.6 + 54.6

=404.2 S cm2 mol−1

Formula of degree of dissociation:

ά = λ°(HCOOH)/ λ°(HCOOH)

ά = 46.1 / 404.2

ά = 0.114

Formula of dissociation constant:

K = (c ά2)/(1 – ά)

Plug the values we get

K = 3.67 × 10–4 mol per liter

Answer 2

Answer:

Hi mate.....✌♥️

Explanation:

Given that

λ0(H+)= 349.6 S cm2 mol–1

λ0(HCOO–) = 54.6 S cm2 mol–1

Concentration ,C = 0.025 mol L-1

λ(HCOOH) = 46.1 S cm2 mol−1

use formula

λ°(HCOOH) = λ0(H+) + λ0(HCOO–)

plug the values we get

λ°(HCOOH) = 0.349.6 + 54.6

=404.2 S cm2 mol−1

Formula of degree of dissociation:

ά = λ°(HCOOH)/ λ°(HCOOH)

ά = 46.1 / 404.2

ά = 0.114

Formula of dissociation constant:

K = (c ά2)/(1 – ά)

Plug the values we get

K = 3.67 × 10–4 mol per liter