Question
Each atom in a crystal of aluminum metal occupies a theoretical cube that is $0.255 \mathrm{nm}$ on a side. If the density of the aluminum crystal is $2.70 \mathrm{~g} / \mathrm{cm}^{3},$ what is the experimental value of Avogadro's number?
Step 1
255 \, \text{nm}$. The volume of a cube is given by the formula $V = a^3$, where $a$ is the length of a side. So, the volume of one atom is $(0.255 \, \text{nm})^3 = 0.0165 \, \text{nm}^3$. Show more…
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