A decinormal solution of `NaCl` has specific conductivity equal to `0.0092`. If ionic conductances of `Na^(+)` and `Cl^(-)` ions at the same temperature are 43.0 and `65.0 ohm^(-1)` respectively, calculate the degree of dissociation of NaCl solution.

Explanation

Equivalent conductance of N/10 NaCl solution
`^^_v=Sp` Conductance`xx`dilution
`=0.0092xxxx10,000=92ohm^(-1)`
`^^_(oo)=lamda_(Na^+)+lamda_(Cr^-)=43.0+65.0=108ohm^(-1)`
Degree of dissociation, `prop=(^^_v)/(^^_(oo))=(92)/(108)=0.85`