The resistance of a conductivity cell when filled with 0.05M solution of an electrolyte x is 100 `Omega` at `40^(@)C`. The same conductivity cell filled with 0.01M solution of electrolyte y has a resistance of `50 Omega`. The conductivity of 0.05 M solution of electrolyte x is `1.0 xx 10^(-4) S cm^(-1)`. Calculate
(i) cell constant
(ii) conductivity of 0.01 M solution
(iii) molar conductivity of 0.01 M solution

Explanation

Given, for electrolyte x, molarity = 0.05 M,
resistance = `100Omega` and
conductivity `= 1.0 xx 10^(-4) S cm^(-1)`
For electrolyte y, molarity = 0.01 M, resistance =` 50 Omega` and conductivity = ?
(i) Cell constant, `G^(**) =` Resistance `( R ) xx `conductivity (k)
`= 100 xx 1.0 xx 10^(-4) = 10^(-2) cm^(-1)`
(ii) Conductivity of solution y,
`k = ("Cell constant")/( "Resistance") = ( 10^(-2))/( 50) = 2xx 10^(-4) S cm^(-1)`
(iii) Molar conductivity ,
`Lambda _(m) = ( k xx 1000)/( "Molarity ") = ( 2 xx 10^(-4) xx 1000)/( 0.01)`
= 20 S `cm^(2) "mol^(-1)`