Question 11: Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant? Chapter 3: Electrochemistry Chemistry Class 12 solutions are developed for assisting understudies with working on their score and increase knowledge of the subjects. Conductivity of 0.00241 M acetic acid is 7.896 ? 10?5 S cm?1. Calculate its molar conductivity and if for acetic acid is 390.5 S cm2 mol?1, what is its dissociation constant? is solved by our expert teachers. You can get ncert solutions and notes for class 12 chapter 3 absolutely free. NCERT Solutions for class 12 Chemistry Chapter 3: Electrochemistry is very essencial for getting good marks in CBSE Board examinations
Question 11: Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?
Answer:Given that , κ = 7.896 × 10−5 S m−1
C=M= 0.00241 mol L−1
The formula of molar conductivity,
Λm = (k × 1000)/M
Plug the value we get
Λm = (7.896 × 10−5 × 1000)/ 0.00241
= 32.76S cm2 mol−1
The formula of degree of dissociation
α = Λm/ Λ°m
Plug the value we get
α = 32.76S/390.5
= 0.084
The formula of dissociation constant
K = Cα/(1 – α)
Plug the values we get
K = 0.00241 × 0.084/(1– 0.084)
= 1.86 × 10−5 mol L−1
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