A cubic metal of molar mass 52 g mol^-1 has unit cell edge 288 pm and density 7.23 g cm^-3. Using N_A = 6.022x10^23 mol^-1, find the number of atoms per unit cell (nearest integer).

JEE Main levelNumerical answerDifficulty 2 of 3ss014

A cubic metal of molar mass $52 g mo l^{- 1}$ has unit cell edge $288 pm$ and density 7.23 $g c m^{- 3}$ . Using $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , find the number of atoms per unit cell (nearest integer).

Numerical answer type, tolerance 0.5.

From

ρ = \frac{ZM}{N _{A} a ^{3}}

Z = \frac{ρ N _{A} a ^{3}}{M}

. With

a^{3} = 2.3888 e - 23 c m^{3}

Z = \frac{7.23 \times 6.022 \times 1 0 ^{23} \times 2.3888 e - 23}{52} = 2.000 \approx 2

, so the lattice is bcc.