How can I find the volume of a single silver atom, along with its radius in angstroms?

Question

1) If I have a cube with a volume of one cm3, and the density of silver is 10.49 g/cm3, how many atoms are in the cube?

I came up with about 5.8e22. Is this right?

2) Since atoms are spherical, they can't occupy all of the space of the cube. The silver atoms pack in the solid in such a way that 74% of the volume of the solid is actually filled with the silver atoms. From this info, what is the volume of a single silver atom?

3) Using the volume from 2), and the formula for the volume of a sphere (four-thirds x pi x radius.cubed), what is the radius in angstroms of a silver atom?

I need these answered really as fast as possible, I don't mean to sound rude about that. Thanks in advance for any help.

Answer 1

1) the mass of the cube is 10.49. Molar mass of silver =107.9

so 10.49/107.9=0.0972mole . you multiply by Avogadro's number and

0.0972*6.022*10^23=5.85*10^22 (so you are not far from the solution

2) the volume of all silver atoms is 0.74cm^

you divide  0.74 by 5.58*10^22=1.264*10^-23cm^3

3) V= (4/3) *pi*r^3  so r^3=3V/4*pi=3.02*10^-24cm^3

r= 1.44*10^-8 cm^3= 1.44 angstrom

Answer 2

1) 5.855 x 10^22 atoms/cm3  (to four sig figs, since the data is four sig figs)  

I NEED PAPER TO WRITE ON for a face centered cubic problem..  and im without a calculator..  and this webpage apparently doesnt allow any sort of helpful formatting