Answer:
There are [tex]6.624 \times 10^{21}\,atoms[/tex] of Germanium in a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si.
Explanation:
The masses of silicon and germanium contained in a cubic centimeter of the germanium-silicon alloy by apply the concepts of mass ([tex]m[/tex]), density ([tex]\rho[/tex]) and volume ([tex]V[/tex]), as well as the mass-mass proportion of Germanium ([tex]x[/tex]):
[tex]m_{Ge} = x \cdot \rho_{Ge}\cdot V_{sample}[/tex]
[tex]m_{Ge} = 0.15\cdot \left(5.32\,\frac{g}{cm^{3}} \right)\cdot (1\,cm^{3})[/tex]
[tex]m_{Ge} = 0.798\,g[/tex]
The amount of moles of Germanium is obtained after dividing previous outcome by its atomic weight. That is to say:
[tex]n = \frac{m_{Ge}}{M_{Ge}}[/tex]
[tex]n = \frac{0.798\,g}{72.64\,\frac{g}{mol} }[/tex]
[tex]n = 0.011\,mol[/tex]
There are 0.011 moles in a cubic centimeter of the germanium-silicon alloy. According to the Law of Avogadro, there are [tex]6.022 \times 10^{23}\,atoms[/tex] in a mole of Germanium. The quantity of atoms in a cubic centimeter is therefore found by simple rule of three:
[tex]y = \frac{0.011\,mol}{1\,mol}\times \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)[/tex]
[tex]y = 6.624 \times 10^{21}\,atoms[/tex]
There are [tex]6.624 \times 10^{21}\,atoms[/tex] of Germanium in a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si.
