Question

An element crystallizes in a cubic close pack structure. The edge of a unit cell is 408 pm and the density of the element is 19.27 g/cm^3. From the atomic mass of the element, the element is :

(a) Ag (b) Au (c) W (d) P

Answer 1

Density of the unit cell = mass of unit cell / volume of unit cell

Volume of unit cell = a³ (where a = edge length)

Mass of unit cell = z x m (z = no. of atom in unit cell and m = mass of one atom)

Also, m = M/ N_A   (Here M = molar mass and N_A = Avogadro's number)

Density = z.M / a³. N_A

For ccp structure, z = 4

Given Density = 19.27 g/cm³ and edge length, a = 408 pm

Volume of unit cell = a³ = (408pm)³ = (408 x 10^-12m) = (408 x 10^-10 cm)³ = 6.79 x 10^-23 cm³

Put all the values in above formula and calculate unknown value of M.

M = d.N_A.a³ / z = 19.27 x 6.022 x 10²³ x 6.79 x 10^-23 / 4 = 196.26 g /mol

Atomic mass of Au = 196.9u

Therefore the element is Au