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.

High-energy and unstable.

There is no reason to think the results are biased or not conducted properly but you should not assume the results are correct just because they were done in a laboratory so the answer would be c. it is likely to be reliable because the results support the conclusion presented.