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The molar conductivity of 0.05 M of solution of an electrolyte is 200 $Ω^{- 1} {cm}^{2} {mol}^{- 1}$ . The resistance offered by a conductivity cell with cell constant (1/3) ${cm}^{- 1}$ would be about

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44.44 $Ω$

33.33 $Ω$

11.11 $Ω$

22.22 $Ω$

answer is C.

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Detailed Solution

$κ = Λ_{c} = (200 S {cm}^{2} {mol}^{- 1}) (0 .05 \times 10^{- 3} mol {cm}^{- 1}) = 0 .01 S {cm}^{- 1}$

$R = \frac{1}{κ} (\frac{ℓ}{A}) = \frac{1}{(0 .01 {S cm}^{- 1})} (\frac{1}{3} {cm}^{- 1}) = 33 .33 Ω$

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