1 mol of electrons passes through each of the solution of ${\mathrm{AgNO}}_3, {\mathrm{CuSO}}_4\mathrm{and} {\mathrm{AlCl}}_3 \mathrm{when} \mathrm{Ag}, \mathrm{Cu} \mathrm{and} \mathrm{Al}$ are deposited. Their molar ratio will be 

${\mathrm{Ag}}^{+}+1{\mathrm{e}}^{-}\to \mathrm{Ag} 1 \mathrm{mole}\to 108\mathrm{gm} \mathrm{for} 1 \mathrm{mole} \mathrm{Ag}$

${\mathrm{Cu}}^{+2}+2\mathrm{e}\to \mathrm{Cu} 2\mathrm{mole}\to 63.5\mathrm{gms} \mathrm{for} \frac{1}{\mathrm{e}}\mathrm{mole} \mathrm{Cu}$$1\mathrm{mole}\to 31.75\mathrm{gm}$

${\mathrm{Al}}^{+3}+3{\mathrm{e}}^{-}\to \mathrm{Al} 3 {\mathrm{mole}}^{-}\to 27 \mathrm{for} 1 \mathrm{mole} \mathrm{Al}\to \frac{27}{3}$

$\mathrm{Ag}:\mathrm{Cu}:\mathrm{Al}=1:\frac{1}{2}:\frac{1}{3}=6:3:2.$