Question

The molar conductivity of 0.025 mol L−1 methanoic acid is 46.1 S cm2 mol−1 .

Answer 1

Given that

λ0(H+)= 349.6 S cm^2 mol–1

λ0(HCOO–) = 54.6 S cm^2 mol–1

Concentration ,C = 0.025 mol L−1

λ(HCOOH) = 46.1 S cm^2 mol−1

use formula

λo(HCOOH)   = λ0(H+)  +   λ0(HCOO–)

plug the values we get

λo(HCOOH)   = 0.349.6 + 54.6

                        =404.2 S cm^2 mol−1

Formula of degree of dissociation:

ά = λo(HCOOH)/ λo(HCOOH)

ά = 46.1 / 404.2

ά = 0.114

Formula of dissociation constant:

K = (c ά^2)/(1 – ά)

Plug the values we get

K=0.025x(0.114)^2/(1-0.114)

K=3.67×10^-4 mol per litre

Answer 2

Answer:

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

 

1−0.1140.025×(0.114)2​

K = 3.67 × 10–4 mol per liter