Question

estimate the number of molecules in 1.00m^3 of air by assuming that it has a density of 1.21kg/m^3 and that is entirely composed of nitrogen molecules

Answer 1

M(N₂) = 28.014 g/mol

ρ(N₂) = 1.21 kg/m³

V(N₂) = 1 m³

$m = ρ \times V \\ m(N_2) = 1.21 \times 1 = 1.21 \;kg = 1210 \;g \\ n=\frac{m}{M} \\ n(N_2) = \frac{1210}{28.014} \\ = 43.19 \;mol \\ N_A = 6.022 \times 10^{23} \\ n = 43.19 \times 6.022 \times 10^{23} \\ = 2.6 \times 10^{25}$

Answer: $2.6 \times 10^{25}$