Question

3. (4 points) A metal crystallizes in a face centered cubic structure and has a density of 11.9 g/cm. If the radius of the metal atom is 138 pm, what is the identity of the metal? P ( i)x (138Yio) 5.9H CX1013 (5.9466X 10 Cm 7.07 X107S 7.07 X103 6.02 2 x 103 -S

Answer 1

Geiven, a meta lizes in a face-centered cubie structure. So, no of atom present in this lattice will be = (% X6) + (/8 X 8). = 4 And we know for this lattice relation between side length and radious of each atom will be ar2 = 4r a= side length [r= radius of atom] or a 40 4X138 = pm [as Y= 138 pm] = 390.3 pm = (390.3 x 10 -10) cm [as tpm = 10-10 om] Now, we know, density of a cell, M = atomic mass z = no of atom in lattice briven, d = 11.9 810m² N = avogadro no Z = 4 La = side length of lattice / N = 6.022 x 1023 a = 390.3 X 18-10 cm d- MZ d = Na3 . dNa3 M = 11.9 X 6.022 x 10 x(390.3 x 10 ) = 106.5 g So, from this calculation we know that the atomic mass an the metal atom is 106.5 g And, atomic mass of Palladium (Pd) will be 106.62 g Hence, it may be palladium (Pd) metal.