At `18^(@)C`, the conductivities at infinite dilution of `NH_(4)Cl, NaOH` and `NaCl` are 129.8, 217.4 and 108.9 mho respectively. If the equivalent conductivity of N/100 solution of `NH_(4)OH` is 9.93 mho, calculate the degree of dissociation of `NH_(4)OH` at this dilution.

Explanation

`wedge_(oo)NH_(4) Cl=lambda_(NH_(4)^(+))=129.8 .......(i)`
`wedge_(oo NaOH)=lambda_(Na^(+))+lambda_(OH^(-))=217.4 .....(ii)`
`wedge_(oo NaCl)=lambda_(Na^(+))+lambda_(Cl^(-))=108.9 ...........(ii)`
Adding eqs. (i) and (ii) and substacting eq. (iii)
`lambda_(NH_(4)^(+))+lambda_(Cl^(-))+lambda_(Na^(+))+lambda_(OH^(-))-lambda_(Na^(+))-lambda_(Cl^(-))`
`lambda_(NH_(4)^(+))+lambda_(CH^(-))=129.8+217.4-108.9`
`wedge_(oo NH_(4)OH)=238.3 mho`
Degree of dissociation `alpha =(wedg_(V))/(wedge_(oo))=(9.93)/(238.3)=0.04167`
Or 4.17 % dissociated