The electrical resistance of a column of `0.05 M 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

`A=pir^(2)=3.14xx0.5^(2)cm^(2)=0.785cm^(2)=0.785xx10^(-4)m^(2)`
`l=50cm=0.5m`
`R=(rhol)/(A)` or `rho=(RA)/(l)=(5.55xx10^(3)Omegaxx0.785cm^(2))/(50cm)=87.135Omega` cm
Conductivity `=k=(1)/(rho)=((1)/(87.135))S" "cm^(-1)`
`=0.01148S" "cm^(-1)`
Molar conductivity `^^_(m)=(kxx1000)/(c)cm^(3)L^(-1)`
`=(0.01148S" "cm^(-1)xx1000cm^(3)L^(-1))/(0.05mol" "L^(-1))`
`=229.6cm^(2)mol^(-1)`
If we want to calculate the values of different quantities in terms of 'm' instead of 'cm'
`rho=(RA)/(l)`
`=(5.55xx10^(3)Omegaxx0.785xx10^(-4)m^(2))/(0.5m)=87.135xx10^(-2)Omega`m
`k=(1)/(rho)=(100)/(87.135)Omega" "m=1.148S" "m^(-1)`
and `^^_(m)=(k)/(c)=(1.148S" "m^(-1))/(50mol" "m^(-3))=229.6xx10^(-4)S" "m^(2)jmol^(-1)`