The electrical resistance of a column of `"0.5 mol.L"^(-1)` NaOH solution of diameter 1 cm and length 50 cm is `5.55xx10^(3) ohm`. Calculate its resistivity, conductivity and molar conductivity.

Explanation

l = 50 cm = 0.5 m and `A=pir^(2)=3.14xx0.5^(2)`
`=0,785cm^(2)=0.785xx10^(-4)m^(2)`.
`R=(rhol)/(A), or rho=(RA)/(l)=(5.55xx10^(3)Omegaxx0.785cm^(2))/(50 cm)=87.135Omega.cm`
`"Conductivity "= kappa =(1)/(rho)=((1)/(87.135))"S.cm"^(-1)="0.01148 S.cm"^(-1)`
`=0.01148xx100S.m^(-1)=1.148S.m^(-1)`
`"Molar conductivity, "Lambda_(m)=(kappa xx 1000)/(C)cm^(2).L^(-1)`
`=(0.01148S.cm^(-1)xx1000cm^(3).L^(-1))/("0.50 mol.L"^(-1))`
`="229.6 S.cm"^(2).mol"^(-1)`
`=229.6S(10^(-2)m)^(2)"mol"^(-1)`
`=229.6xx10^(-4)"S.m"^(2)."mol"^(-1)`