Question

A drop ($0.05$ mL) of $12MHCl$ is spread over a thin sheet of aluminium foil (thickness $0.10$ mm and density of Al=$2.70$g/mL). Assuming whole of the HCl is used to dissolve Al, the maximum area of the hole produced in the foil will be (write it as $0.6=6$ (only one digit)) :

Answer 1

Meq. of Al = Meq. of HCl
=$12\times 0.05=0.6$
$\therefore$ Mass of Al =$\frac{0.6\times 9}{1000}=0.0054g$
Volume of Al foil =$\frac{0.0054}{2.7}$ mL or ${cm}^3=0.002{cm}^3$
Now , $Area\times thickness=Volume$
$\therefore Area=\frac{0.002}{0.01}=0.2{cm}^2$ (thickness=0.01 cm)
So, answer is 2.