Question #183476

A hypothetical alloy has a face-centered cubic unit cell. Its density is 10,470 kg/m^{3}_{,} and its atomic weight is 106.8 g/mole.

a) What is the atomic radius of the alloy in cm?

b) What is the volume of the single atom of the alloy in cm^{3}?

Expert's answer

Convert the density into "g\/cm3"

D="10470kg\/m3"

D therefore= "10.47g\/cm3"

"d=(z\u00d7M)\/(a3\u00d7Na)"

For a FCC "Z=4"

Therefore,

10.47g/cm3= "(4\u00d7106.8g\/mol)\u00f7(a3\u00d76.022\u00d710^23)\n\na3=6.776\u00d710^-22."

But volume"= l3"

L= "3\u221avol"

"L=3\u221a6.776\u00d710^-22\n\nL=8.79\u00d710^-8cm"

"=8.79\u00d710^-8\u00d7(10^10\/10^2cm)"

=8.79∆

But, "C^2=a^2+a^2\n,C^2=2a^2"

"C^2=2\u00d7(8.79)^2"

"C^=\u221a154.53"

"C=12.43\u2206"

And,

"r=c\/4"

r=12.43/4

Ans for part( a)

"r=3.12" ans.

(B)

Vol of a single atom "V=4\/3\u03c0r^3"

"V=4\/3\u00d722\/7\u00d73.12^3"

"V=127.27cm^3" Ans.

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Assignment Expert28.04.21, 15:05Dear klein lyse, You're welcome. We are glad to be helpful. If you liked our service please press like-button beside answer field. Thank you!

klein lyse27.04.21, 17:01thank you so much!

klein lyse27.04.21, 16:59cobalt crystallizes in a HCP unit cell. its atomic radius is 0.1253 nm. its density is 8900 kg/m^3. using this information, estemate avogadro's number

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