Question

Find the mass of $1.5055\times {10}^{25}$ number of molecule of aluminium oxide.

Answer 1

Molucullar weight of $A{l}_2{O}_3=102g$

$6.022\times {10}^{23}$ molecules are present in 102 g $A{l}_2{O}_3$

So $1.5055\times {10}^{25}$ molecules will present in $\frac{102}{6.022\times {10}^{23}}\times 1.5055\times {10}^{25}$ g $A{l}_2{O}_3$

$=2.55\times {10}^3$ g