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Electrochemistry

Question

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250 cm³ solution containing 1 g of Ca(NO₃)₂ in conductivity cell offered a resistance of 48 ohms. The cell constant of conductivity of the solution is 0.56 cm^-1. Find the molar conductivity.

Solution

{}^{n}{Ca} ({NO}_{3})_{2} = \frac{1 g}{164 g {mol}^{- 1}} = 0.0061 mol V = 250 {cm}^{3} = 0.250 L

∴

M = \frac{n}{V} = \frac{0.0061}{0.250} = 0.0244 M K = \frac{Cell constant}{Resistance} = \frac{0.56}{48} = 1.167 \times 10^{- 2} {ohm}^{- 1} {cm}^{- 1}

\land_{m} = \frac{10^{3} K}{M} = \frac{10^{3} \times 1.167 \times 10^{- 2}}{0.0244} \land_{m} = 478.28 s {cm}^{2} {mol}^{- 1} .

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