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.

Verified Answer :

Hagrid

here's my best guess

the volume of the unit cell is (38510^-12)^3=5.706610^-29 m^3

multiply by density to get mass
mass = (7 g/cm^3)(100^3 cm^3 / 1^3 m^3) 5.706610^-29 m^3= 3.9946610^-22 g

covert to moles
3.9946610^-22 g 1 mol / 239.82 g = 1.6657 10^-24 mol

convert to number of units

1.6657 10^-24 mol 6.2310^23 units/mol = 1.04

385 pm = 3.8510^(-8) cm

The volume of the unit cell is the cube of that, which is 5.7110^(-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/(5.7110^(-23)), where x is the mass of the unit cell. Solving for x, you get 410^(-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.