The electrical resistance of a column of 0.05 M KOH solution of diameter 1 cm and length 45.5 cm is `4.55 xx 10^(3)` ohm. Calculate its molar conductivity.

Explanation

Given diameter of column = 1 cm
Length of the column (l) = 45.5 cm
Area of cross-section of the column `=pi xx r^(2) = 3.14 xx (0.5)^(2) = 0.785 cm^(2)`
We have the relation:
`R = rho 1/A` or `1/rho = 1/R xx 1/A` or `k =1/R xx 1/A`
Substituting the values, we have
Conductivity, `k =1/(4.55 xx 10^(3) "ohm") xx (45.5 cm)/(0.785 cm^(2)) = 0.0127 S cm^(-1)`
Molar conductivity, `Lambda_(m) = k xx 1000/C`
`=0.0127 S cm^(-1) xx 1000/(0.05 "moles" cm^(-3)) = 254 S cm^(2) mol^(-1)`.