Question

13. (13 pts.) Calculated the th below. (Assuming ro-2=0.140 nm. Calculated the theoretical density of Mgo which has a NaCl structure shown ning to 0.140 nm, I'Mg2+ =0.072 nm, Ame=24.305 g/mol, A.-16 g/mol). Show all calculations steps to receive full credit. 1.56x102 hardnes s n=6 1.50 Santa n (Eute) - v NA 2 Mg2+

Answer 1

Density of MgO will be given as: = (ZO.MO + ZMg.MMg) / (NA x V)

Where ZO = ZMg = no. of atoms per unit cell = 4 (Because MgO follows NaCl structure and in an NaCl structure no. of atoms per unit cell = 4)

MMg = molecular mass of Mg = 24.305 g/mol and MO = molecular mass of oxygen = 16 g/mol

NA = Avogadro's number = 6.022 x 1023 mol-1

V = volume of a unit cell = (edge length)3 = a3

As MgO follows NaCl therefore it is an FCC cubic ctructure. For FCC structure, relation between edge length and radius of atoms are:

a = 2(rMg2+ + rO2-) = 2(0.072 nm + 0.140 nm) = 0.424 nm = 4.24 x 10-8 cm

So, V = a3 = (4.24 x 10-8)3 = 76.22 x 10-24 cm3

Therfore density = [4(24.305 g/mol) + 4(16 g/mol)] / (6.022 x 1023 mol-1 x 76.22 x 10-24 cm3)

= 161.22 / 45.899 = 3.51 g/cm3