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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.
Solution
`k= 0.0058 "ohm"^-1 "cm"^-1`
∧m `= ("k"xx1000)/("Molarity")`
`=(0.0058xx1000)/(5/208xx1000/1000)`
`=(0.0058xx1000xx28)/5`
`=(208xx5.8)/5`
`= 208xx1.16`
`= 241.28 "ohm"^-1 "cm"^2 "mol"^-1`
∧eq = ∧m `xx"Eq.wt.of BaCI"_2/"MoI.wt.of BaCI"_2`
= ∧m `xx(208/2)/208`
∧eq =` (241.28xx1)/2`
`= 120.64 "ohm"^-1 "cm"^2 "eq"^-1`
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