The conductivity of a solution containing 1.0 g of anhydrous BaCl, in 200 cm of the solution has been found to 0.0058 S `cm^(-1)`. Calculate the molar conductivity and equivalent conductivity of the solution.

Answer

`Lambda_(m) = 241.67 S cm^(2) mol^(-1) , Lambda = 120 .83 S cm^(2) "equiv"^(-1)`

Explanation

Molarity of solution = `(1//208 xx 1000)/(200) = 0.0240 M`
`Lambda_(m) = (0.0058 xx 1000)/(0.0240) = 241.67 ohm^(-1) cm^(2) mol^(-1)`
Normality of solution = `(1//104 xx 1000)/(200) = 0.0480 N`
`Lambda = (0.0058 xx 1000)/(0.0480) = 120.83 ohm^(-1) cm^(2) "equiv"^(-1)`