Class 12 Chemistry - Chapter Electrochemistry NCERT Solutions | Conductivity of 0.00241 M acetic acid is

Welcome to the NCERT Solutions for Class 12th Chemistry - Chapter Electrochemistry. This page offers a step-by-step solution to the specific question from Exercise 2, Question 11: conductivity of 0 00241 m acetic acid is 7 896 ti....
Question 11

Conductivity of 0.00241 M acetic acid is 7.896 × 10 - 5 S cm - 1. Calculate its molar conductivity and if Amº for acetic acid is 390.5 S cm2 mol - 1, what is its dissociation constant?

 

Answer

Given,K = 7.896 × 10 - 5 S m - 1

c = 0.00241 mol L - 1

Then, molar conductivity, Am = K/c

= 32.76S cm2 mol - 1

Again, Amº = 390.5 S cm2 mol - 1

Now,

= 0.084

Dissociation constant,

= (0.00241 mol L-1)(0.084)2  / (1-0.084)

= 1.86 × 10 - 5 mol L - 1

More Questions From Class 12 Chemistry - Chapter Electrochemistry

Popular Questions of Class 12 Chemistry

Write a Comment: