Given that the density of tlcl(s) is 7.00 g/cm3 and that the length of an edge of a unit cell is 385 pm, determine how many formula units of tlcl there are in a unit cell.

Here's my best guess
the volume of the unit cell is (385*10^-12)^3=5.7066*10^-29 m^3
multiply by density to get mass mass = (7 g/cm^3)*(100^3 cm^3 / 1^3 m^3) * 5.7066*10^-29 m^3= 3.99466*10^-22 g
covert to moles 3.99466*10^-22 g * 1 mol / 239.82 g = 1.6657 *10^-24 mol
convert to number of units

1.6657 *10^-24 mol * 6.23*10^23 units/mol = 1.04

385 pm = 3.85*10^(-8) cm

The volume of the unit cell is the cube of that, which is 5.71*10^(-23) cm^3. Since the ratio of mass to volume (i.e. the density) must be the same no matter what amount of TlCl you have, you can say:
7 = x/(5.71*10^(-23)), where x is the mass of the unit cell. Solving for x, you get 4*10^(-22) g.

The mass of a molecule of TlCl is 240 amu, which in grams is 4*10^(-22) g. The mass of the unit cell and the mass of a molecule of TlCl is the same. Therefore there is one formula unit of TlCl per unit cell.