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Conductance and Conductivity

The specific ...

Question

The specific conductance of a solution containing 5 g of anhydrous $B a C l_{2}$ (mol. wt. = 208) in 1000 $c m^{3}$ of a solution is found to be 0.0058 $o h m^{- 1} c m^{- 1}$ . Calculate the molar and equivalent conductivity of the solution.

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Solution

Given parameters:
$κ = 0.0058 Ω^{- 1} {c m}^{- 1}, m = 5 g, M = 208 g / m o l, V = 1000 m L$
Molar conductivity:
$Λ_{m} = \frac{κ \times 1000}{M o l a r i t y} M o l a r i t y = \frac{m}{M} \times \frac{1000}{V} = \frac{5}{208} \times \frac{1000}{1000} = 0.024 M Λ_{m} = \frac{κ \times 1000}{M o l a r i t y} = \frac{0.0058 \times 1000}{0.024} = 241.7 Ω^{- 1} {c m}^{2} {m o l}^{- 1}$
Equivalent conductivity:
$Λ_{e} = \frac{κ \times 1000}{N o r m a l i t y} N o r m a l i t y = \frac{M o l a r i t y}{g r a m e q u i v a l e n t} = \frac{0.024}{2} = 0.012 N Λ_{e} = \frac{0.0058 \times 1000}{0.012} = 483.33 Ω^{- 1} {c m}^{2} {(g e q)}^{- 1}$

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