$$4.5$$ $$g$$ of aluminium ( atomic mass $$27u$$  ) is deposited at cathode from $$A{l^{3 + }}$$  solution by a certain quantity of electric charge. The volume of hydrogen produced at $$STP$$  from $${H^ + }\,ions$$  in solution by the same quantity of electric charge will be

Solution :
$$\eqalign{ & {\text{From second law of Faraday}} \cr & \,\,\,\,\,\,\frac{{{m_{Al}}}}{{{m_H}}} = \frac{{{E_{Al}}}}{{{E_H}}} \cr & \,\,\,\,\,\,\frac{{4.5}}{{{m_H}}} = \frac{{\frac{{27}}{3}}}{1} \cr & {\text{or}}\,\,\,{m_H} = 0.5\,g \cr & \because \,\,{\text{Volume of}}\,\,2\,g\,{H_2}\,{\text{at}}\,STP = 22.4\,L \cr & \therefore \,\,{\text{Volume of}}\,\,0.5\,g\,{H_2}\,{\text{at}}\,STP \cr & \,\,\,\,\,\, = \frac{{22.4 \times 0.5}}{2}L \cr & \,\,\,\,\,\, = 5.6\,L \cr} $$