Question

Given molar conductivity of an infinite dilution: `wedge_(m)^(@)` for `Ba(OH)_(2)=517.6Omega^(-1)cm^(2)mol^(-1)`.
`wedge_(m)^(@)` for `BaCl_(2)=240.6Omega^(-1)cm^(2)mol^(-1),wedge_(m)^(@)` for `NH_(4)Cl=129.8Omega^(-1)cm^(2)mol^(-1)`. Calculate `wedge_(m)^(@)` for `NH_(4)OH`.

Answer 1

Correct Answer - `268.3Omega^(-1)cm^(2)mol^(-1)`
`wedge_(m)^(@)(Ba(OH)_(2))=lamda_(Ba^(2+))^(@)+2lamda_(OH^(-))^(@)` . . . (i)
`wedge_(m)^(@)(BaCl_(2))=lamda_(Ba^(2+))^(@)+2lamda_(Cl^(-))^(@)` . . . (ii)
`wedge_(m)^(@)(NH_(4)Cl)=lamda_(NH_(4)^(+))^(@)+lamda_(Cl^(-))^(@)` . . . (iii)
`wedge_(m)^(@)(NH_(4)OH)=lamda_(NH_(4)^(+))^(@)+lamda_(OH^(-))^(@)+lamda_(OH^(-))^(@)=(1)/(2)eqn.(i)+eqn.(iii)-(1)/(2)`
`=(1)/(2)xx517.6+129.8-(1)/(2)xx240.6=268.3Omega^(-1)cm^(2)mol^(-1)`