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

1 Answer
Sep 28, 2016

Answer:

#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.