Question

The conductivity of `0.001028M` acetic acid is `4.95xx10^(-5)Scm^(-1)`. Calculate its dissociation constant if `Lambda_(m)^(0)` for acetic acid is `390.5Scm^(2)mol^(-1)`.

Answer 1

We know,
`Lambda_(m)=kxx(1000)/(M)`.....(i)
Given : `"k=4.95 x 10"^(-5)"S cm"^(-1),M=0.001028`
`therefore"From"(i)Lambda_(m)=4.95xx10^(-5)xx(1000)/(0.001028)`
`="48.15 ohm"^(-1)cm^(2)mol^(-1)`
Degree of dissociation,