Question

An element X (atomic weight = 24 gm/mol)forms a face centered cubic lattice. If the edgelength of the lattice is 4 x 10- cm and theobserved density is 2.40 x 10 kg/m”, then thepercentage occupancy of lattice points byelement X is : (Use N. = 6 x 1023):​

Answer 1

Hello Dear,

◆ Answer -

Percentage occupancy = 96 %

◆ Explanation -

# Given -

M = 24 g/mol = 24×10^-3 kg/mol

a = 4×10^-8 cm = 4×10^-10 m

d' = 2.4×10^3 kg/m^3

# Solution -

For FCC lattice, density of unit cell is calculated as -

d = Z.M / NA.a^3

d = (4 × 24×10^-3) / [6×10^23 × (4×10^-10)^3]

d = 2.5×10^3 kg/m^3

Percentage occupancy in lattice structure will be -

% Occupancy = observed density / calculated density × 100

% Occupancy = 2.4×10^3 / 2.5×10^3 × 100

% Occupancy = 96 %

Hence, percentage occupancy lattice of element X is 96 %.

Best luck dear...