1.

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»


SOLUTION :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)`


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