Specific conductance of 0.1 M CH(3)COOH ...

Specific conductance of 0.1 M `CH_(3)COOH` at `25^(@)` C is `3.9xx10^(-4) " ohm"^(-1) "cm"^(-1)`
If `lambda^(infty)(H_(3)O^(+))` and `lambda^(infty)(CH_(3)COO^(-))` at `25^(@)`C are `349.0 and41.0 " ohm"^(-)"cm"^(2) "mol"^(-1)` respectively degree of ionisation of `CH_(3)COOH` at the given concentration is

`1.0 %`

`4.0 %`

`5.0%`

`2.0%`

Text Solution

Verified by Experts

The correct Answer is:

`alpha=(kxx1000//conc.)/(lambda_(CH_(3)COO^(-))^(prop)+lambda_(H_(3)O^(+))^(prop))`

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