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Question
The conductivity of 0.01 mol/dm3 aqueous acetic acid at 300 K is 19.5×10−5ohm−1cm−1 and the limiting molar conductivity of acetic acid at the same temperature is 390 ohm−1cm2mol−1. The degree of dissociation of acetic acid is:
The conductivity of 0.01 mol/dm3 aqueous acetic acid at 300 K is 19.5×10−5ohm−1cm−1 and the limiting molar conductivity of acetic acid at the same temperature is 390 ohm−1cm2mol−1. The degree of dissociation of acetic acid is:
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Solution
The correct option is B 0.05
Data:-
Concentration(C)=0.01moldm−3=10molm−3.Conductivity(K)=19.5×10−5Scm−1=19.5×10−3Sm−1
Limiting molar conductivity(∧∘m)=390Scm2mol−1=390×10−4Sm2mol−1
Solution:-(i) Molar Conductivity∧m=KC=19.5×10−3Sm−110molm−3∧m=19.5×10−4Sm2mol−1.
(ii) Degree of dissociation(α)=∧m∧∘m=19.5×10−4Sm2mol−1390×10−4Sm2mol−1∴α=0.05
Data:-
Concentration(C)=0.01moldm−3=10molm−3.
Conductivity(K)=19.5×10−5Scm−1=19.5×10−3Sm−1
Limiting molar conductivity(∧∘m)=390Scm2mol−1=390×10−4Sm2mol−1
Solution:-
(i) Molar Conductivity∧m=KC=19.5×10−3Sm−110molm−3
∧m=19.5×10−4Sm2mol−1.
(ii) Degree of dissociation(α)=∧m∧∘m=19.5×10−4Sm2mol−1390×10−4Sm2mol−1
∴α=0.05
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acetic acid is 390.5 S cm2
mol−1,
what is its dissociation constant?