The electrical resistance of a column of 0.05 mol `L^(-1)` NaOH solution of diameter 1 cm and length 50 cm is `5.55 xx 10^3` ohm. Calculate its resistivity, conductivity and molar conductivity.

Explanation

`R=5.55xx10^(3)` ohm `l=50cm`
Diameter = 1 cm Radius = `(1)/(2)` cm
Area of cross section = `pir^(2)=3.14xx((1)/(2))^(2)=0.785cm^(2)`
Cell constant (G) = `(l)/(a)=(50)/(0.785)=63.69cm^(-1)`
Resistivity (P) = `R(l)/(a)=(5.55xx10^(3))/(63.69)=87.13Omegacm`
Conductivity `(K)=(1)/(P)=(1)/(87.13)=1.14xx10^(-2)Scm^(-)`
Molar conductivity = `(Kxx1000)/(M)=(1.14xx10^(-2)xx1000)/(0.05)=229.6Scm^(2)mol^(-1)`