The limiting molar ionic conductivity of Na+ is 50.1 S cm^2 mol^-1. Taking F = 96500 C mol^-1, the ionic mobility of Na+ at infinite dilution is

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The limiting molar ionic conductivity of $Na X^{+}$ is $50.1 S c m^{2} mo l^{- 1}$ . Taking $F = 96500 C mo l^{- 1}$ , the ionic mobility of $Na X^{+}$ at infinite dilution is

For an ion of charge number

z

λ^{\circ} = z u F

; for

Na X^{+}

z = 1

u = λ^{\circ} / F

u = \frac{50.1 S c m ^{2} mo l ^{- 1}}{96500 C mo l ^{- 1}} = 5.19 \times 1 0^{- 4} c m^{2} V^{- 1} s^{- 1}

(since

S c m^{2} C^{- 1} = c m^{2} V^{- 1} s^{- 1}

). Multiplying by

F

instead gives

4.83 \times 1 0^{6}