Question

If a metal forms a FCC lattice with unit edge length 500 pm. Calculate the density of the metal if its atomic mass is 110.

(a) 2923 kg/m^3

(b) 5846 kg/m^3

(c) 8768 kg/m^3

(d) 1750 kg/m^3

I had been asked this question by my school teacher while I was bunking the class.

Question is from Solid State in section Solid State of Chemistry – Class 12

Answer 1

Right option is (b) 5846 kg/m^3

The best explanation: Given,

Edge length (a) = 500 pm = 500 x 10^-12 m

Atomic mass (M) = 110 g/mole = 110 x 10^-3 kg/mole

Avogadro’s number (NA) = 6.022 x 10^23/mole

z = 4 atoms/cell

The density, d of a metal is given as d=\(\frac{zM}{a^3N_A}\)

On substitution, d=\(\frac{4 \times 110 \times 10^{-3}}{(500 \times 10^{-12})^3 \times 6.022 \times 10^{23}}\)=5846 kg/m^3.