Chemistry - Electrochemistry Question with Solution | TestHub

${\Lambda }_m^{\infty }$$({\mathrm{AlCl}}_3)$ = ${\lambda }_m(\mathrm{A}{\mathrm{l}}^{3+})$ + 3${\lambda }_m(\mathrm{C}{\mathrm{l}}^{-})$.

${\lambda }_m(\mathrm{A}{\mathrm{l}}^{3+})$ = 3${\lambda }_{eq}(\mathrm{A}{\mathrm{l}}^{3+})$ and ${\lambda }_m(\mathrm{C}{\mathrm{l}}^{-})$ = ${\lambda }_{eq}(\mathrm{C}{\mathrm{l}}^{-})$.

So ${\Lambda }_m^{\infty }$ = 3${\lambda }_{eq}(\mathrm{A}{\mathrm{l}}^{3+})$ + 3${\lambda }_{eq}(\mathrm{C}{\mathrm{l}}^{-})$ = 3(${\lambda }_{eq}(\mathrm{A}{\mathrm{l}}^{3+})$ + ${\lambda }_{eq}(\mathrm{C}{\mathrm{l}}^{-})$).

Answer: (C).