The electrical resistance of a column `0.05 mol L^(-1)` NaOH solution of diameter 1 cm and length 50 cm is `5.55 xx10^(3)` ohm. Calculate its resistivity, conductivity and molar conductivity.

Explanation

A ` = pi r^(2) = 3.14 xx 0.5^(2) cm^(2) = 0.785 cm^(2) = 0.785 xx 10^(4) m`
` l = 50 cm = 0.5 m`
`R = (rho l)/(A)` or `rho = (RA)/(l)= (5.55 xx 10^(3) Omega xx 0.785 cm^(2))/(50 cm) = 87.135Omega cm`
Conductivity `= k = (1)/(rho) = (1)/((87.135)) S Cm^(-1) `
` = 0.01148 S cm^(-1)`
Molar conductivity , `^^_(m) = (k xx 1000)/(c) cm^(3)L^(-1)`
` = (0.01148 S cm^(-1) xx 1000 cm^(3) L^(-3))/(0.05 mol L^(-1))`
= `229.6 S cm^(2) mol^(-1)`
If we want to calculate the value of different quantities in terms of 'm' instead of cm .
`rho = (RA)/(l)`
` = (5.55 xx 10^(3) Omega xx0.785 xx 10^(-4) m^(2))/(0.5 m)`
` = 87.135 xx 10^(-2) Omega m`
` k = (1)/(p) = (100)/(87.135) Omega m = 1.148 S m^(-1)`
and `^^_(m) = (k)/(c) = (1.148 S m^(-1))/(50 mol m^(-3)) = 229.6 xx 10^(-4) S m^(2) mol^(-1) `