The resistance offered by 0.1 M solution of acetic acid when placed in a conductivity cell of length 1.0 cm and area of cross section $3.0 {cm}^{2}$ is found to be $7 \times 10^{3} Ω$ . The molar conductivity of the solution is

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$4.76 S {cm}^{2} {mol}^{- 1}$

$4.76 \times 10^{- 4} S {cm}^{2} {mol}^{- 1}$

$0.476 S {cm}^{2} {mol}^{- 1}$

$4.76 \times 10^{- 3} S {cm}^{2} {mol}^{- 1}$

answer is C.

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Detailed Solution

$κ = \frac{1}{R} (\frac{l}{A}) = (\frac{1}{7 \times 10^{3} Ω}) (\frac{1.0 cm}{3.0 {cm}^{2}}) = 4.76 \times 10^{- 5} S {cm}^{- 1}$

$Λ = \frac{κ}{c} = \frac{4.76 \times 10^{- 5} S {cm}^{- 1}}{0.1 mol {dm}^{- 3}} = 4.76 \times 10^{- 4} S {dm}^{3} {cm}^{- 1} {mol}^{- 1} = 4.76 \times 10^{- 4} S (10 cm)^{3} {cm}^{- 1} {mol}^{- 1} = 0.476 S {cm}^{2} {mol}^{- 1}$

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