## Materials Science and Engineering: An Introduction

$N = 3.30 \times 10^{28} atoms/m^{3}$
Required: Number of atoms per cubic meter in Pb Solution: $p= 11.35 g/cm^{3}$ $A_{Pb} = 207.2 g/mol$ Using Equation 4.2: $N = \frac{N_{Pb}p_{Pb}}{A_{Pb}} =\frac{(6.022 \times 10^{23} atoms/mol)(11.35 g/cm^{3})}{207.2 g/mol} = 3.30 \times 10^{22} atoms/cm^{3} = 3.30 \times 10^{28} atoms/m^{3}$