# Let's Review: The Tools of Quantitative Chemistry - Study Questions: 61

a) $1.143\times10^{21}$ atoms, $76.4\%$ of the volume. b) 4

#### Work Step by Step

a) Number of atoms: $cube's\ mass/atom's\ mass=0.1206/(1.055\times10^{-22})=1.143\times10^{21}$ Volume of one atom ($r=128\times10^{-12}\ m$): $V_a=\frac43\pi r^3=8.78\times10^{-30}\ m^3=8.78\times10^{-24}\ cm^3$ Volume of that number of atoms: $V_m=1.00\times10^{-2}\ cm^3$ Volume of the cube ($L=0.236\ cm$): $V_t=L^3=1.31\times10^{-2}\ cm^3$ Fraction of the cube which is occupied by atoms: $V_m/V_t\times100\%=76.4\%$ Spheres touching each other leave room "unutilized" due to their geometry. b) Volume of the cube ($L=361.47\times10^{-10}\ cm$): $V=L^3=4.7230\times10^{-23}\ cm^3$ With the density of 8.960 g/cm³, the mass of the cube is $m_c=4.232\times10^{-22}\ g$ Since the mass of an atom is $1.055\times10^{-22}$, there are $4.011\approx 4$ atoms in the cube.

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