Question #268330

Gold has a face centered cubic unit cell and has the density 𝟏𝟗. 𝟑 𝒈 𝒄𝒎−𝟑. Calculate the

length along an edge. (𝐴𝑢 = 197 𝑔 𝑚𝑜𝑙−1)


1
Expert's answer
2021-11-21T17:16:15-0500

FCC type gold contain atom

18×8+6×12=4\frac{1}{8}\times8+6\times\frac{1}{2}=4

Mass of unit cell of FCC type

4×1976.023×1023=1.30×10214\times\frac{197}{6.023\times10^{23}}=1.30\times10^{-21} g

Density of gold =19.3g/cm3

Density=massofunitcellvolumeofunitcell\frac{mass of unit cell }{volume of unit cell}

Volume of unit cell =massofunitcelldensityofgold\frac{mass of unit cell }{density of gold}


V=1.30×102119.3=6.73×1023cm3V=\frac{1.30\times10^{-21}}{19.3}=6.73\times10^{-23}cm^3

V=a3=6.73×1023cm3V=a^3=6.73\times10^{-23}cm^3

a=4.06×108cma=4.06\times10^{-8}cm

For FCC type unit cell

a=8ra=\sqrt{8}r


r=a8=4.06×1088=1.43×108cmr=\frac{a}{\sqrt8}=\frac{4.06\times10^{-8}}{\sqrt8}=1.43\times10^{-8}cm


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS