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Q.

The molar ionic conductance at infinite dilution of ${Ag}^{+} is 61.92 \times 10^{- 4} S {mol}^{- 1} m^{2}$ at $25^{°} C$ the ionic mobility of ${Ag}^{+}$ will be

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a

$6.4 \times 10^{- 8}$

b

$6.192 \times 10^{- 4}$

c

6.192

d

$3.2 \times 10^{- 4}$

answer is A.

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Detailed Solution

$μ_{{Ag}^{+}} = \frac{μ_{{Ag}^{+}}}{F} = \frac{61.92 \times 10^{- 4}}{96.500}$

$= 6.4 \times 10^{- 8}$

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The molar ionic conductance at infinite dilution of Ag+ is 61.92×10-4S mol-1m2at 25°C the ionic mobility of Ag+ will be