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Electrochemistry

Question

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The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55 x 10³ ohm. Calculate its resistivity, conductivity and molar conductivity.

Solution

l = 50 cm A = {πr}^{2} = 3.14 \times (0.5)^{2} = 0.785 cm Cell constant = \frac{l}{A} = \frac{50}{0.785} = 63.694 {cm}^{- 1}

(i)

Resistivity = \frac{R}{Cell constant} = \frac{5.55 \times 10^{3}}{63.694} = 87.135 Ω cm .

(ii)

Conductivity = \frac{1}{resistivity} = \frac{1}{87.135} = 1.148 \times 10^{- 2} s {cm}^{- 1}

(iii)

Molar conductivity \land_{m} = \frac{10^{3} K}{M} = \frac{10^{3} \times 1.148 \times 10^{- 2}}{0.05} \land_{m} = 229.6 s {cm}^{2} {mol}^{- 1}

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