The correct order of conductivity of ions in water is:

$C s^{+} > R b^{+} > K^{+} > N a^{+}$

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$K^{+} > N a^{+} > C s^{+} > R b^{+}$

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$R b^{+} > N a^{+} > K^{+} > L i^{+}$

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$N a^{+} > K^{+} > R b^{+} > C s^{+}$

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Solution

The correct option is A
$C s^{+} > R b^{+} > K^{+} > N a^{+}$

Answer: (a)
Hydration of $M^{+}$ alkali metals decrease down the group.
Hence, sodium is highly hydrated and cesium is less hydrated.
Since, $C s^{+} (a q)$ has a lower hydrated radius it is free to move so its electrical conductivity is higher.

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Calculate the molar conductivity of water at infinite dilution (in Sm $^{2}$ mol $^{- 1}$ ).

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A conducts water and ions whereas B conducts food.

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The correct order of conductivity of ions in water is:

The correct option is A Cs+>Rb+>K+>Na+ Answer: (a) Hydration of M+ alkali metals decrease down the group. Hence, sodium is highly hydrated and cesium is less hydrated. Since, Cs+(aq) has a lower hydrated radius it is free to move so its electrical conductivity is higher.