Respuesta :
Explanation:
As gold sheet will be in the form of a rectangle so, volume of the gold sheet will be as follows.
Volume = length × breadth × height
As given values are length is 68.1 nm, breadth or width is 28.8 nm, and height is 7.78 nm.
Putting these values into the above formula to calculate the volume as follows.
Volume = length × breadth × height
= 68.1 nm × 28.8 nm × 7.78 nm
= 15258.75 [tex]nm^{3}[/tex]
As 1 nm = [tex]10^{-9} m[/tex]. Hence, converting volume into meter as follows.
15258.75 [tex]nm^{3} \times \frac{10{-27}m^{3}}{1 nm^{3}}[/tex]
= [tex]1.52 \times 10_{-23} m^{3}[/tex]
It is given that unit cell edge length is 4.08 [tex]^{o}A[/tex] = [tex]4.08 \times 10^{-10} m[/tex] m
Volume of a unit cell = [tex]a^{3}[/tex] = [tex](4.08 \times 10^{-10}m)^{3}[/tex]
= [tex]6.79 \times 10^{-29} m^{3}[/tex]
So, number of unit cells in the gold sheet = [tex]\frac{\text{volume of gold sheet}}{\text{volume of unit cell}}[/tex]
= [tex]\frac{1.52 \times 10_{-23}}{6.79 \times 10^{-29} m^{3}}[/tex]
= [tex]0.223 \times 10^{6}[/tex]
= [tex]2.23 \times 10^{5}[/tex]
A face-centerd cubic cell contain 4 atoms of gold.
Hence, number of atoms in the sheet are calculated as follows.
[tex]4 \times 2.23 \times 10^{5}[/tex]
= [tex]8.92 \times 10^{5}[/tex] gold atoms
As 1 mol contains [tex]6.023 \times 10^{23}[/tex] atoms. So, number of moles in [tex]8.92 \times 10^{5}[/tex] gold atoms will be calculated as follows.
No. of moles of gold atoms in the sheet = [tex]\frac{8.92 \times 10^{5}}{6.023 \times 10^{23}}[/tex]
= [tex]1.48 \times 10^{-18}[/tex] mol
Thus, we can conclude that given fcc arrangement has [tex]1.48 \times 10^{-18}[/tex] moles of gold atoms.
