Metallic rhodium crystallizes in a face-centered cubic lattice, with one Rh atom

Chemistry
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Metallic rhodium crystallizes in a face-centered cubic lattice, with one Rh atom per lattice point. If the metallic radius of Rh is 134 pm, what is the volume of the unit cell in pm3 and in cm3?

 pm3
 cm3
Jun 2nd, 2015

Please let me know if you need any clarification. I'm always happy to answer your questions.

When you hear face-centered cubic (FCC), the edge length is 3*root(2) times the radius. 

This means the answer to your first question is radius=length/(3root2)=91.5pm 

The edge length on your second problem is length=radius*(3root2)=568pm. So the volume is 568^3=1.83x10^8 pm^3 

pm^3 to cm^3, divide it by 10^30,

 So it's 1.83x10^-22 cm^3

1.83x10^8 pm^3--------------------cm^3

1.83x10^-22 cm^3

1.83x10^8 pm^3-


Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 2nd, 2015

That was so wrong heres the answer

For a face-centered cubic lattice with one atom per lattice point, there are4 atoms per unit cell. 
The atoms are in contact along a face diagonal and 4r = (2)1/2a, where r is the metallic radius and a is the edge length of the cell.

Thus: a = r / 0.354 = 134 pm /0.354 = 379 pm

For a cubic cell the volume is the edge length cubed.
V = (379 pm)3 = 5.44E7 pm3

1 pm = 10-12m = 10-10cm

1 pm3 = 10-30cm3

5.44E7 pm3 
  =  5.44E-23 cm3


Jun 2nd, 2015

okay dear, iam so sorry that i did not get it right..........................................

Jun 2nd, 2015

you see how its done?

Jun 2nd, 2015

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