wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A weak monobasic acid is 5% dissociated in 0.01 mol dm3 solution. The limiting molar conductivity at infinite dilution is 4.00×102ohm1m2mol1. Calculate the conductivity of a 0.05 mol dm3 solution of the acid.

Open in App
Solution

Dissociation constant of acid Ka=Cα2
=0.01×(0.05)2
=2.5×105
Ka=Cα2
2.5×105=0.05×α2
α=0.0223
We know that, α=cmm
0.0223=cm4×102
cm=8.92×104ohm1cm2mol1.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Resistivity and Conductivity
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon