A bcc metal has atomic mass 56 g mol^-1. Using N_A = 6.022x10^23 mol^-1, the number of unit cells present in 10.0 g of the metal is

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A bcc metal has atomic mass $56 g mo l^{- 1}$ . Using $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , the number of unit cells present in $10.0 g$ of the metal is

Atoms

= \frac{10.0}{56} \times 6.022 \times 1 0^{23} = 1.08 \times 1 0^{23}

. A bcc cell holds 2 atoms, so cells

= 1.08 \times 1 0^{23} /2 = 5.38 \times 1 0^{22}