Chemistry - Electrochemistry Question with Solution | TestHub

${\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}^{\mathrm{\infty }}$for$\mathrm{N}{\mathrm{H}}_4\mathrm{C}\mathrm{l}$can be find out by the relation given below
${\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}^{\mathrm{o}}$or${{\mathrm{\Lambda }}^{\mathrm{\infty }}}_{\left(\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}\right)}={{\mathrm{\Lambda }}^{\mathrm{\infty }}}_{\mathrm{e}\mathrm{q}\left(\mathrm{N}{\mathrm{H}}_4\mathrm{C}\mathrm{l}\right)}+{{\mathrm{\Lambda }}^{\mathrm{\infty }}}_{\mathrm{e}\mathrm{q}\left(\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}\right)}-{{\mathrm{\Lambda }}^{\mathrm{\infty }}}_{\mathrm{e}\mathrm{q}\left(\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}\right)}$
$=(149.74+248.1-126.4)$
$=271.44{\text{ohm}}^{-1}{\text{cm}}^2{\text{eq}}^{-1}$