The specific conductivity of a solution containing `1.0g` of anhydrous `BaCl_(2)` in `200 cm^(3)` of the solution has been found to be `0.0058 S cm^(-1)`. Calculate the molar and equivalent conductivity of the solution. Molecular wt. of `BaCl_(2) = 208`[mu implies `lambda_(m)]`

Explanation

Molarity, of `BaCI_(2) = (1 xx 1000)/(208 xx 200) = 0.024 M`
Also, Normality of `BaCI_(2) = 0.024 xx 2xx = 0.048 M ( :' N = M xx V.f)`
Now, `^^_(m) = k xx (1000)/(C_(M)) = (0.0058 xx 1000)/(0.024) = 241.67 S cm^(2) mol^(-1)`
Also, `^^_(eq) = k xx (1000)/(C_(N)) = (0.0058 xx 1000)/(0.048) = 120.83 S cm^(2) "equivalent"^(-1)`