Question

The specific conductivity of a solution containing 5 g of anhydrous BaCl₂ (mol. wt. = 208) in 1000 cm³ of a solution is found to be 0.0058 ohm⁻¹ cm⁻¹. Calculate the molar and equivalent conductivity of the solution.

Answer 1

κ = 0.0058 ohm⁻¹ cm⁻¹

Λ_m = `(kappa xx 1000)/("Molarity")`

= `(0.0058 xx 1000)/(5/208 xx 1000/1000)`

= `(0.0058 xx 1000 xx 208)/5`

= `(208 xx 5.8)/5`

= 208 × 1.16

= 241.28 ohm⁻¹ cm² mol⁻¹

Λ_eq = `Lambda_m xx ("Equivalent weight of BaCl"_2)/("Molecular weight of BaCl"_2)`

= `(241.28 xx 1)/2`

= 120.64 ohm⁻¹ cm² eq⁻¹