206. The conductivity of 0.025 mol L- methanoic acid is46.1 s cm2 mol-

1.	206. The conductivity of 0.025 mol L- methanoic acid is46.1 s cm2 mol-!. Calculate the degree ofdissociation and dissociation constant. (Given, a349.6 S cmmol-= 54.6 S cm mol-(HCOO)020
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 K = 3.67 × 10–4 mol per liter

206. The conductivity of 0.025 mol L- methanoic acid is46.1 s cm2 mol-!. Calculate the degree ofdissociation and dissociation constant. (Given, a349.6 S cmmol-= 54.6 S cm mol-(HCOO)020

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

K = 3.67 × 10–4 mol per liter