Question

5.238 x 10^24 molecules of aluminum bromide to grams

Answer 1

2319.698 g

Procedure

Use the Avogadro's number and then the molecular weight to convert into grams. The formula for aluminum bromide is AlBr₃.

`5.238*10^(24)\text{ molecules AlBr}_3\frac{1\text{ mole}}{6.022*10^(23)}\frac{266.69\text{ g}}{1\text{ mole}}=2319.698\text{ grams}`