What volume of aluminum has the same number of atoms as 10 cm3 of mercury?

In terms of calculating the macroscopic system state variables are going to be used.

The $\rho$is one of the densities of state variables and is defined through the ratio of another two-state variable and it is provided through the equation ($18.1$) in the form

$$\begin{gathered} \rho =\frac{M}{V} \\ V=\frac{M}{\rho } \end{gathered}$$ $\left(1\right)$

Then, the atoms' number N is provided through equation (18.4) in the form

$$\begin{gathered} N=\frac{M}{m} \\ M=mN \end{gathered}$$

In this case, N is the atoms' number and m is molecular mass. The M's expression is going to be used in equation (1)

$$\begin{gathered} V=\frac{mN}{p}\left(2\right) \\ N=\frac{pV}{m}\left(3\right) \end{gathered}$$

The mercury's molecular mass is $m_{Hg}$=$200$u and density is pHg=$13600$kg/$m^3$.

The equation (3) is going to be used in order to get $N_{Hg}$through:

localid="1651396197699" $$\begin{gathered} \mathrm{N}=\frac{\mathrm{pV}}{\mathrm{m}} \\ =\frac{(13600\mathrm{kg}/{\mathrm{m}}^3)(10\times {10}^{-6}{\mathrm{m}}^3)}{200\times 1.66\times {10}^{-27}\mathrm{kg}} \\ =4.09\times {10}^{23}\mathrm{atoms} \end{gathered}$$

The aluminium has the same numbers as the subject of atoms. Therefore equation (2) is going to be used to get the volume.

Aluminium's molecular mass is $m_{AI}=27u$

density is $p_{AI}=2700kg/m^3$

$V=\frac{mN}{p}$

localid="1651396213148" $$\begin{gathered} =\frac{(27\times 1.66\times {10}^{-27}\mathrm{kg})(4.09\times {10}^{23}\mathrm{atoms})}{2700\mathrm{kg}/{\mathrm{m}}^3} \\ =6.8\times {10}^{-6}{\mathrm{m}}^3 \\ =6.8\mathrm{c}.{\mathrm{m}}^3 \end{gathered}$$

Therefore, the result is$6.8c{m}^3$