Answer:
[tex]\rho = 4.63 g/mL[/tex]
Explanation:
Volume of each drum in which the Uranium Hexafluoride is contained is given as
[tex]V = 1.62 \times 10^6 L[/tex]
[tex]V = 1.62 \times 10^9 mL[/tex]
now we know that mass of the uranium is given as
[tex]m = 2.25 \times 10^8 kg[/tex]
now total volume of all 30 containers is given as
[tex]V = 30 (1.62 \times 10^9)[/tex]
[tex]V = 4.86 \times 10^{10} mL[/tex]
[tex]m = 2.25 \times 10^{11} g[/tex]
so now the density is given as
[tex]\rho = \frac{m}{V}[/tex]
[tex]\rho = \frac{2.25 \times 10^{11}}{4.86 \times 10^{10}}[/tex]
[tex]\rho = 4.63 g/mL[/tex]
