To calculate the mass of BeI₂, we first need to convert it to number of moles using the Avogadro's number:
[tex]n_{BeI_2}=\frac{N_{BeI_2}}{N_A}=\frac{3.6\times10^{24}}{6.02\times10^{23}mol^{-1}}=0.5980\ldots\times10^1mol=5.980\ldots mol[/tex]Now we need to calculate the molar mass of BeI₂ using the molar masses of the atoms Be and I, which we can consult on a periodic table:
[tex]\begin{gathered} M_{BeI_2}=1\cdot M_{Be}+2\cdot M_I \\ M_{BeI_2}=(1\cdot9.01+2\cdot126.90)g\/mol \\ M_{BeI_2}=(9.01+253.80)g\/mol \\ M_{BeI_2}=262.81g\/mol \end{gathered}[/tex]And now we can calculate the mass:
[tex]\begin{gathered} M_{BeI_{2}}=\frac{m_{BeI_2}}{n_{BeI_{2}}} \\ m_{BeI_2}=n_{BeI_2}\cdot M_{BeI_{2}} \\ m_{BeI_2}=5.980\ldots mol\cdot262.81g\/mol \\ m_{BeI_2}\approx1571.62g \end{gathered}[/tex]Which corresponds to alternative A.