The electical 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

Area `(a)=pi r^(2)=3.14xx((1cm)/(2))^(2) `
`=0.785 cm^(2)`
`=0.785xx10^(-4)m^(2)`
`l=50 cm=0.5m`
`R=(rhol)/(a)` or `rho=(Ra)/(l)=(5.55xx10^(3)ohmxx0.785cm^(2))/(50cm)`
`=87.135ohm cm`
Conductivity `(k)=(1)/(rho)=((1)/(87.135))S cm^(-1)`
`=0.01148 S cm^(-1)`
Molar conductivity `(wedge_(m))=(kxx1000)/(M)cm^(3)L^(-1)`
`=(0.01148Scm^(-1)xx1000cm^(3)L^(-1))/(0.05mol L^(-1))`
`=229.6 S cm^(2) mol^(-1)`
Alternatively
The value of different quantities in `SI` units `(i.e.,` in terms of `"m"` instead of `"cm")`
`rho=(Ra)/(l)=(5.55xx10^(3)ohm xx 0.785 xx 10^(-4)m^(2))/(0.5m)`
` =87.135xx10^(-2)ohm m`
`k=(1)/(rho)=(100)/(87.135)ohm m=1.148 S m^(-1)`
and `wedge _(m)=(k)/(c)=(1.148Sm^(-1))/(50 mol m^(-3))=229.6xx10^(-4)S m^(2) mol ^(-1)`