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Electrochemistry

Question

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The measured resistance of a conductance cell containing 7.5 x 10^–3 M solution of KCl at 25° C was 1005 ohms. Calculate (a) specific conductance (b) molar conductance of the solution cell constant = 1.25 cm^–1.

Solution

We have given
Resistance = 1005ohm
cell constant =1.25 cm^-1thus
Specific conductance,

k = \frac{1}{R} \times cell constant = \frac{1}{1005} \times 1.25 \Rightarrow 1.244 \times 10^{- 3} S {cm}^{- 1} Λ_{m} = \frac{1000 K}{M} \Rightarrow \frac{1000 \times 1.244 \times 10^{- 3}}{7.5 \times 10^{- 3}} S {cm}^{2} {mol}^{- 1} \Rightarrow 165 s {cm}^{2} {mol}^{- 1} .

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