Chemistry Liquids and Solids Finding unit cells

Question

The density of Solid Ag is 10.5 g.cm^3. How many atoms are present per cubic centimeter of Ag?

* __ atoms/cm³

As a solid Ag adopts a face centered cubic unit cell.

How many Unit cells are present per cubic centimeter of Ag?

*__ unit cells/cm³

What is the volume of a Unit cell of this metal?

*__ cm³

What is the edge length of a Unit cell of Ag?

*__ cm

Answer 1

Gold's amu is 196.96655u, multiply that by 1.66E^-24 to convert to grams...

 

One atom of gold weighs 3.26964473E^-22g

 

Divide the density 10.5 g.cm³ by the weight of one atom (3.26964473E-22g) to get atoms/cm³

 

10.5 / 3.26... = 3.211358073E²² atoms/cm³

 

A face centered cubic unit cell contains 4 atoms (6 halves and 8 eighths), so divide that last answer by 4 to get the number of unit cells.

 

3.21... / 4 = 8.028395183E²¹ unit cells/cm³

 

Take the reciprocal of that to get cm³/unit cell, so 1 / 8.02... = 1.24557894E^-22 cm³/unit cell

 

Then you're dealing with a cube's volume (V = L*L*L), and you want just the length, so take the cube root of your last answer and you get 4.994098296E^-8 cm

 

I think that's right, it's been a while since I did chemistry. This guy does a good video of a very similar problem. https://www.youtube.com/watch?v=qe8GjH-N1FY He converts to mols at the beginning and uses avagadro's number. That would work too.