The specific conductivity of a solution containing `1.0g` of anhydrous `BaCI_(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 `BaCI_(2) = 208`[mu implies `lambda_(m)]`

Explanation

We are given: Conductivity `(kappa)=0.0058" S "cm^(-1)`
Molar mass of `BaCl_(2)=137+2xx35.5=208" g "mol^(-1)`
As 1 gram of `BaCl_(2)` is present in 200 `cm^(3)` of the solution, therefore, molar concentration (c)
`=(1)/(208)xx(1)/(200)xx1000" mol "L^(-1)=0.0240" mol "L^(-1)`
`therefore`Molar conductivity, `wedge_(m)=(kappaxx1000)/(c_(m))=(0.0058" S "cm^(-1)xx1000cm^(3)L^(-1))/(0.0240" mol "L^(-1))=241.67" S "cm^(2)mol^(-1)`
Further in case of `BaCl_(2)`, equivalent weight`=("Mol. wt")/(2)=(208)/(2)=104`
`therefore` Concentration of the solution in gram equivalent per litre `(c_(eq))=(1)/(104)xx(1)/(200)xx1000=0.0480`
`therefore`Equivalent conductivity, `wedge_(eq)=(kappaxx1000)/(c_(eq))=(0.0058" S "cm^(-1)xx1000cm^(3)L^(-1))/(0.0480" g "eqL^(-1))=120.83" S "cm^(2)eq^(-1)`.