A 0.005 cm thick coating of copper is deposited on a plate of 0.5 ${{m}^{2}}$ total area. Calculate the number of copper atoms deposited on the plate (density of copper = 7.2 g $c{{m}^{-3}}$, atomic mass = 63.5).
Area of plate = 0.5 ${{m}^{2}}$ = 0.5 x ${{10}^{4}}$ $c{{m}^{2}}$
Thickness of coating = 0.005 cm
Volume of copper deposited = 0.5 x ${{10}^{4}}$ x 0.005 = 25 $c{{m}^{3}}$
Mass of copper deposited = 25 X 7.2 = 180 g
Now, 63.5 g of copper contains atoms = 6.022 X ${{10}^{23}}$
.•. 180 g of copper will contain atoms = 6.022 x ${{10}^{23}}$/63.5 x 180 = 1.71X ${{10}^{24}}$