Solution to Set 4

Problem 1

Cu, density $\rho =8.885{g \over cm^{3}}$, atomic mass $m_{a}=63.57amu$, fcc structure

a. In $1m^{3}$, the number of moles is ${8.885g \over 1cm^{3}}{1cm^{3} \over 5\times 10^{6}m^{3}}{1mol \over 63.55g}=1.40\times 10^{5}{mol \over m^{3}}$

b. Atoms in $1m^{3}$ ${1.40\times 10^{5}mol \over 1m^{3}}{6.022\times 10^{23} \over 1mol}=8.43\times 10^{28}atoms$