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Electrochemistry

Question

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Calculate the resistance offered by 0.5 M CH₃COOH solution when its molar conductivity. is 7.4 Ω^–1 cm² mol^–1 and the cell constant is 0.037 cm^–1.

Solution

We have gien the molar conductivity 7.4Ω^–1 cm² mol^–1
cell constant 0.037cm^-1thus by using formula

k = \frac{1}{R} \times (\frac{l}{A})

and therefore

R = \frac{1}{k} \times (\frac{l}{A}) k = Λ_{m} \times C = 7.4 Ω^{- 1} {cm}^{2} {mol}^{- 1} \times 0.5 mol / L = 7.4 Ω^{- 1} \times \frac{0.05 mol}{1000 {cm}^{3}} = 3.7 \times 10^{- 4} Ω^{- 1} {cm}^{- 1} .

∴

R = \frac{1}{k} . \frac{l}{a}

= \frac{1}{3.7 \times 10^{- 4} Ω^{- 1} {cm}^{- 1}} \times 0.037 {cm}^{- 1} = 100 Ω .

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