The elctrical resistance of a column of `0.05 M N aOH ` solution of diameter `1 cm` and length `50 cm` is `5.55 xx 10^(3)ohm`. Calculate its resisteivity , conductivity, and molar conductivity.

Explanation

(i) Calculation of resistivity. Electrical resistance of the solution, `R=5.55xx10^(3)Omega`
Area of cross-section of the column `(a)=pir^(2)=3.14xx((1)/(2))^(2)cm^(2)=0.785cm^(2)`
Length of the column `(l)=50cm`
Applying the formula `R=rho(l)/(a)`
`rho=(Rxxa)/(l)=((5.55xx10^(3)Omega)(0.785cm^(2)))/(50cm)=87.135Omegacm` i.e., Resistivity `(rho)=87.135Omega` cm
(ii) Calculation of conductivity. Conductivity `(kappa)=(1)/(rho)=(1)/(87.135Omega" cm")=0.01148" S "cm^(-1)`
(iii) Calculation of molar conductivity
Molar conductivity `(wedge_(m))=(kappaxx1000)/("Molarity")=((0.01148" S "cm^(-1))(1000cm^(3)L^(-1)))/((0.05" mol "L^(-1)))=229.6" S "cm^(2)mol^(-1)`