Question

A drop ($0.05$ mL) of $12.0MHCl$ is spread over a sheet of thin aluminium foil. Assuming that the acid dissolves through the foil, the area is $x\times {10}^{-1}c{m}^2$ of the hole produced, then $x$ is :

(Density of $Al=2.70gc{m}^{-3}$; thickness of the foil $=0.10$ mm)

Answer 1

$0.05$ mL of $12.0MHCl$ corresponds to $0.6$ millimoles of $HCl$.

It will react with $0.2$ millimoles of $Al$ or $27\times 0.2=5.4$ mg of $Al$.

The volume of $Al$ will be $\frac{5.4}{1000\times 2.70}=0.002c{m}^3$

The area of the hole will be $\frac{0.002}{0.01}=0.2c{m}^2=2\times {10}^{-1}$