A ccp lattice of anions contains $0.35$ mol of anions. Cations occupy one-fourth of the tetrahedral voids and one-half of the octahedral voids. Using $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , find the number of cations in units of $1 0^{22}$ (nearest integer).

Numerical answer type, answer in x 10^22, tolerance 0.5.

Answer: 21 x 10^22

Per anion: tetrahedral voids

2 \times \frac{1}{4} = 0.5

and octahedral voids

1 \times \frac{1}{2} = 0.5

, so 1 cation per anion. Cations

= 0.35 \times 6.022 \times 1 0^{23} = 21.08 \times 1 0^{22} \approx 21 \times 1 0^{22}