Test Index

NCERT Class XII Chemistry
Chapter - Electrochemistry
Questions with Solutions

© examsnet.com
Question : 27
Total: 32
The molar conductivity of 0.025 mol L‌–1 methanoic acid is 46.1 S cm‌2 mol‌–1. Calculate its degree of dissociation and dissociation constant. Given λ(H‌+ ) = 349.6 S cm‌2 mol‌–1 and l(HCOO‌– ) = 54.6 S cm‌2 mol‌–1
Solution:  
Λm0( HCOOH)=λ0( H+)+λ0(HCOO−)
=349.6+54.6=404.2 S cm2 mol−1
α=
λm
λm0
=
46.1
404.2
=0.114
⇒α=11.4%
Ka=
Cα2
1−α

=
0.025×(0.114)2
1−0.114

=
0.025×0.114×0.114
0.886

=3.67×10−4
© examsnet.com
Go to Question:
Prev Question
Next Question