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
`wedge_(v)=Sp." Conductivitiy "xx "dilution"`
`=0.0092 xx 10,000=92" ohm"^(-1)`
`wedge_(oo)=lambda_(Na^(+))+lambda_(Cl^(-))`
=43.0+65.0=108 `ohm^(-1)`
Degree of dissociation, `alpha=(wedge_(v))/(wedge_(oo))=(92)/(108)=0.85`