SELECT * FROM question_mgmt as q WHERE id=1804 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=58 and ex_no='2' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th chemistry Electrochemistry conductivity of 0 00241 m acetic acid is 7 896 ti

Question:

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


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