The conductivity of 0.1 M NaOH is $0.022 S {cm}^{- 1}$ . When equal volume of 0.1 M HCl is added to this solution, the conduction of the solution becomes $0.00632 S {cm}^{- 1}$ . The molar conductivity of sodium chloride in the resultant solution is

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$63.2 S {cm}^{2} {mol}^{- 1}$

$126.4 S {cm}^{2} {mol}^{- 1}$

$189.6 S {cm}^{2} {mol}^{- 1}$

$112.2 S {cm}^{2} {mol}^{- 1}$

answer is A.

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Detailed Solution

The concentration of NaCl in the resultant solution will be 0.05 M. Hence,

$Λ = \frac{κ}{c} = \frac{0.00632 S {cm}^{- 1}}{0.05 mol {dm}^{- 3}} = 0.1264 S {cm}^{- 1} {mol}^{- 1} {dm}^{3} = 0.1264 S {cm}^{- 1} {mol}^{- 1} (10 cm)^{3} = 126.4 S {cm}^{2} {mol}^{- 1}$

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