Answer
$6.8cm^3$
Work Step by Step
The required volume of aluminum can be determined as follows:
As given that the number of atoms is the same for both aluminum, therefore
$\frac{m_{Al}}{M_{Al}}=n=\frac{m_{Hg}}{M_{Hg}}$
$\implies \frac{\rho_{Al}V_{Al}}{M_{Al}}=\frac{\rho_{Hg}V_{Hg}}{M_{Hg}}$
This can be rearranged as:
$V_{Al}=\frac{\rho_{Hg}}{\rho_{Al}}\frac{M_{Al}}{M_{Hg}}V_{Hg}$
We plug in the known values to obtain:
$V_{Al}=(\frac{13600}{2700})(\frac{27}{200})(10)$
This simplifies to:
$V_{Al}=6.8cm^3$
