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

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Answer 1 · 2016-09-28T01:46:01

$2.228 × 10^"23"$ molecules.

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.

How many molecules of N_2O_3N_2O_3 are present in 28.3 g of N_2O_3N_2O_3?