The resistance of a solution is $200 Ω$ when placed in a conductivity cell of cell constant $0.4 {cm}^{- 1} .$ If the molar conductivity of the solution is $20 S {cm}^{2} {mol}^{- 1}$ , the concentration of the solution will be

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0.01 M

0.1 M

$10^{- 3} M$

$10^{- 4} M$

answer is B.

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Detailed Solution

$κ = \frac{K_{cell}}{R} = \frac{0.4 {cm}^{- 1}}{200 Ω} = 2 \times 10^{- 3} S {cm}^{- 1}$

$c = \frac{κ}{Λ_{m}} = \frac{2 \times 10^{- 3} S {cm}^{- 1}}{20.0 S {cm}^{2} {mol}^{- 1}} = 1.0 \times 10^{- 4} mol {cm}^{- 3} = 1.0 \times 10^{- 4} mol {(10^{- 1} dm)}^{- 3} = 0.1 mol {dm}^{- 3}$

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