# Let's Review: The Tools of Quantitative Chemistry - Study Questions - Page 43d: 43

a) $1.78\times10^{-22}cm^3$ b) $3.86\times10^{-22}\ g$ c) $9.65\times10^{-23}\ g$

#### Work Step by Step

a) The volume of a cube is given by $V=L^3$, with a side of $L=0.563\ nm$: $V=0.563^3=0.178\ nm^3$ $0.178\ nm^3\times\left(\dfrac{1 \times 10^2\ cm}{1\times10^9\ nm}\right)^3=1.78\times10^{-22}cm^3$ b) Given the density ($\rho=2.17\ g/cm^3$), the mass is: $m = 2.17 \times 1.78\times10^{-22} = 3.86\times10^{-22}\ g$ c) Since this mass is equivalent to four NaCl units: $m_u=3.86\times10^{-22}\div4=9.65\times10^{-23}\ g$

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