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Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If 0 m ? for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant?

Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If 0 m ? for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant?

Grade:8

2 Answers

Gaurav
askIITians Faculty 164 Points
9 years ago


Given that ,
κ= 7.896 × 10−5S m−1
C=M= 0.00241 molL−1
Theformula ofmolar conductivity,
Λm= (k× 1000)/M
Plug the value we get
Λm= (7.896 × 10−5×1000)/0.00241
= 32.76S cm2mol−1
Theformula ofdegree of dissociation
α= Λm/ Λom
Plug the value we get
α= 32.76S/390.5= 0.084
Theformula ofdissociation constant
K = Cα/(1 – α)
Plug the values we get
K = 0.00241 × 0.084/(1– 0.084)= 1.86 × 10−5mol L−1

Krish Gupta
askIITians Faculty 82 Points
3 years ago
LEt me give a step by step and detailed explanation of the answer
Given that ,
κ= 7.896 × 10−5S m−1
C=M= 0.00241 molL−1
Theformula ofmolar conductivity,
Λm= (k× 1000)/M
Plug the value we get
Λm= (7.896 × 10−5×1000)/0.00241
= 32.76S cm2mol−1
Theformula ofdegree of dissociation
α= Λm/ Λom
Plug the value we get
α= 32.76S/390.5= 0.084
Theformula ofdissociation constant
K = Cα/(1 – α)
Plug the values we get
K = 0.00241 × 0.084/(1– 0.084)= 1.86 × 10−5mol L−1
I hope this works fine,
please approve it.

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