Question

conductivity of 0.00241 M acetic acid is 7.896 into 10 raise to power 5 centimetre inverse calculate its molar conductivity if they not came for acetic acid is 390. 5 cm square mole inverse what is its dissociation constant​

Answer 1

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