1.

Molar conductivities of Li^(-),Na^(+),K^(+) and Rb^(+) ions in aqueous solutions are in the following order.

Answer»

`Li^(+) GT Na^(+) = K^(+) lt Rb^(+)`
`Li^(+) gt Na^(+) gt K^(+) = Rb^(+)`
`Rb^(+) gt K^(+) gt Na^(+) gt Li^(+)`
`Li^(+) gt Rb^(+) gt K^(+) gt Na^(+)`

Solution :The smaller the size of the ion, the greater is the degree of hydration. Therefore, the extent of hydration decreases from `Li^(+) to Rb^(+)`So, hydrated ionic radii decreases in the order: `Li^(+) gt Na^(+) gt K^(+) gt Rb^(+)`
As a result, hydrated Li ion being largest in ionic size, has lowest mobility in AQUEOUS solution. On the other hand, hydrated `Rb^(+)` ion being smallest in size has the HIGHEST mobility in aqueous solution. Higher the ionic mobility, greater is the molar conductivity. Thus, the DECREASING order of molar conductivity is,`Rb^(+) gt K^(+) gt Na^(+) gt Li^(+)`


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