Answer
a) $3.3\times10^{22}\ atoms$
b) $1.82\times10^{-23}\ cm^3$
$0.193\ nm$
Work Step by Step
a) Volume of the cube: $1.000\ cm^3$
Mass of the cube: $11.35\ g$
Atomic weight of lead: $207.2\ g/mol$
Number of moles: $11.35\ g\div207.2\ g/mol=5.48\times10^{-2}\ mol$
$5.48\times10^{-2}\ mol\times6.022\times10^{23}\ atoms/mol=3.3\times10^{22}\ atoms$
b) 60% of the volume: $0.6\ cm^3$
Atomic volume: $0.6\ cm^3\div3.3\times10^{22}\ atoms= 1.82\times10^{-23}\ cm^3$
Radius: $\sqrt[3]{3\times1.82\times10^{-23}\ cm^3\div4\pi}=1.93\times10^{-8}\ cm=0.193\ nm$
