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

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

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

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

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detailed solution

Correct option is A

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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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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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