Calculating resistivity, conductivity and motor conductivity : The electrical resistance of a column of `0.05` mol `L^(-1) NaOH` solution of diameter `1 cm` and length `50 cm` is `55 xx 10^(3) ohm`. Calculate its resistivity, conductivity and molar conductivity
Strategy: Assuming the column of solution to be cylindrical determine the area of cross-section, `A`. Using it along with the length of column, calculate resistivity. Reciprocal of resistivity yields conductivity. When we divide conductivity by molar concentration, we get molar conductivity.

Explanation

Considering cylindrical column of solution, we have
`A = pir^(2) = ((22)/(7)) (0.5 cm)^(2) = 0.785 cm^(2)`
`= 0.785 (10^(-2)m)6(2)`
`= 0.785 xx 10^(-4) m^(2)` ltbr. `l = 50 cm = 0.5 m`
Resistivity can be calculated by Equation (3.16)
`rho = (RA)/(1) = ((5.55 xx 10^(3)Omega)(0.785 xx 10^(-4) m^(2)))/((0.5 m))`
`= 87.135 xx 10^(-2)Omegam`
conductivity can be calculated by Equation (3.18):
`kappa = (1)/(rho) = (1)/(87.135 xx 10^(-2) Omegam)`
`= (100)/(87.135)Omega^(-1) m^(-1)`
`= 1.147 S m^(-1)`
Molar conductivity can be calculated by equation (3.23)
`Lambda_(m) = (K)/(C) = (1.147 Sm^(-1))/(((1000 L)/(m^(3)))(0.05 (mol)/(L))`
`= (1.147 S m^(-1))/(50 mol m^(-3))`
`= 0.02294 S m^(2)mol^(-1)`
`= 229.4 xx 10^(-4) Sm^(2)mol^(-1)`