If nearly $${10^5}C$$  liberate $$1\,g$$  equivalent of aluminium, then the amount of aluminium (equivalent weight 9) deposited through electrolysis in $$20\,min$$  by a current of $$50\,A$$  will be

Solution :
Total charge flowing through electrolyte,
$$\eqalign{ & q = it \cr & = 50 \times 20 \times 60\,\,\left[ {_{t\, = 20 \times 60\,s}^{{\text{as}}\,i = \,50\,A}} \right] \cr & = 6 \times {10^4}C \cr & \therefore {10^5}C\,{\text{release}} = 9{\text{ }}g{\text{ of }}Al \cr & \therefore 6 \times {10^4}C\,{\text{would}}\,{\text{release}} \cr & = \frac{{9 \times 6 \times {{10}^4}}}{{{{10}^5}}}g \cr & = 5.4g\,{\text{of}}\,Al \cr} $$