A gold sample contains $90\%$of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?

If a gold sample contains $90\%$of gold and the rest copper then, the number of atoms of gold present in one gram of this sample of gold is $2.77\times {10}^{21}$.

Let the total mass of the given sample is $100$g.
Mass of gold $= 90$g
The remaining mass is of copper.
⇒ So, the Mass of copper $= (100-90)$
$= 10$g
One gram of gold sample contains gold
$\frac{90}{100} =0.9$g of gold
As we know,
$\Rightarrow \mathrm{Number} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{gold} =\frac{\mathrm{Mass} \mathrm{of} \mathrm{gold}}{\mathrm{Molecular} \mathrm{mass} \mathrm{of} \mathrm{gold}}$
$\Rightarrow \frac{0.9}{197}=0.0046$
One mole of gold or any substance contains Avogadro's number of atoms $=6.022\times {10}^{23}$
So, $0.0046$moles of gold will contain$= 0.0046\times 6.022\times {10}^{23}$ $=2.77\times {10}^{21}$