Bạn tham khảo:
$a/$
$\%Cl^{37}=\frac{450}{450+1350}.100\%=25\%$
$\%Cl^{35}=100\%-25\%=75\%$
$Cl=\frac{450.37+1350.35}{450+1350}=35,5\%$
$b/$
$\%Cl_{(AlCl_3)}=\frac{35,3.3}{133,5}.100\%=79,8\%$
$c/$
$n_{Cl_2}=0,01(mol)$
$n_{Cl^{35}}=\frac{0,01.75}{100}=0,0075(mol)$
$Cl^{35}=0,0075.6,02.10^{23}=4,515.10^{21}$ ntử