Question

How many molecules of `N_2O_3` are present in 28.3 g of `N_2O_3`?

Answer 1

`2.228 × 10^"23"` molecules.

Explanation:

1 mole contains `6.022140857 × 10^"23"` (Avogadro's number)

The molecular weight of `N_2O_3` is: 76.01 g/mol

You have 28.3 g, so you have 28.3 / 76.01 = 0.37 mole

Hence you have `0.37 × 6.022140857 × 10^"23" = 2.228 × 10^"23"`

You have `2.228 × 10^"23"` molecules.