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Molecular Mass

The transport...

Question

The transport no. of $A g^{+}$ ion in $A g N O_{3}$ is 0.48. The equivalent conductivity of $A g N O_{3}$ at infinite dilution is $120 S c m^{- 1} e q^{- 1}$ .

Calculate the limiting ionic conductances and ionic mobilities at infinite dilution:

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Solution

Given equivalent conductivity is $120$

Transport number $t_{A g}^{+}$ is 0.48

For $(N O_{3})^{-}$ the transport number is $1 - 0.48 = 0.52$

equivalent conductivity for $((N O_{3})^{-}$ is $120 \times 0.52$ = $62.4$

Ionic mobility $\frac{62.4}{96500} = 6.467 \times 10^{- 4}$

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