A conductivity cell whose cell constant is 3.0 `cm^(-1)` is filled with 0.1 M solution of a weak acid. Its resistance is found to be 3000 `Omega`. If `wedge_(m)^(infty)= 400 Omega^(-1) cm^(2) mol^(-1)` , the degree of dissociation of the weak acid is about

Options

A. `0.015`
B. `0.025`
C. `0.035`
D. `0.055`

Answer

B

Explanation

`k= (K)/(R )= (3.0 cm^(-1))/(3000Omega)= 0.001S cm^(-1) and Lamda_(m)=(k)/(c )= (0.001 S cm^(-1))/(0.1 xx 10^(-3) mol cm^(-3))= 10Scm^(2) mol^(-1)`
Degree of dissociation, `alpha= (Lamda_(m))/(Lamda_(m)^(oo))= (10)/(400)= 0.025`