Question

The resistance of a conductivity cell filled with `0.1 M KCl` solution is `100 Omega`. If `R` of the same cell when filled with `0.02 M KCl` solution is `520 Omega`, calculate the conductivity and molar conductivity of `0.02 M KCl` solution. The conductivity of `0.1 M KCl `solution is `1.29 S m^(-1)`.
A. `1.24 xx 10^(-4) S cm^(2) mol^(-1)`
B. `1240 xx 10^(-4) Scm^(2) mol^(-1)`
C. `1.24 xx 10^(4) Scm^(2) mol^(-1)`
D. `12.4 xx 10^(4) S cm^(2) mol^(-1)`

Answer 1

Correct Answer - A
`R = rho = (1)/(A),R` = resistance, `rho =` resistivity
`(1)/(A)=` cell constant
`R = (1)/(K)(1)/(A) K = (1)/(rho) =` conductivity
`100 = (1)/(1.29)(1)/(A)`
`:. (1)/(A) = 129 m^(-1)`
For second case
`K = (1)/(R) xx (1)/(A)`
`=(1)/(280) xx 129 Sm^(-1)` (as same cell is used)
Molar conductance `Delta_(m) = (K)/(C)`,
(Where C is concentration in `(mol)/(m^(3)))`
`:. Delta_(m) = (129)/(520 xx 20) S m^(2) mol^(-1)` ltbr. `= 1.24 xx 10^(-2) S m^(2) mol^(-1)`
`=1.24 xx 10^(-4) S m^(2) mol^(-1)`