Resistance of a conductivity cell filled with 0.02 M KCl solution is 520 `Omega`. Calculate the conductivity and molar conductivity of that solution.
[Cell constant of the cell `= 1.29 cm^(-1)`].

Explanation

Given `c=0.02 "mol L"^(-1)`
`R=520 Omega`
`C^(**) = 1.29 cm^(-1)`
Conductivity (K) `=(G^(**))/(R )= (1.29 cm^(-1))/(520)`
Molar conductivity `(wedge_(m))=K xx 1000 cm^(3)L^(-1)//`molarity
`=(0.248 xx 10^(-2)"Scm"^(-1) xx 1000 cm^(3)L^(-1))/(0.02"mol L"^(-1))`
`=124 "Scm"^(2)"mol"^(-1)`