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
M(N2) = 28.014 g/mol
ρ(N2) = 1.21 kg/m3
V(N2) = 1 m3
"m = \u03c1 \\times V \\\\\n\nm(N_2) = 1.21 \\times 1 = 1.21 \\;kg = 1210 \\;g \\\\\n\nn=\\frac{m}{M} \\\\\n\nn(N_2) = \\frac{1210}{28.014} \\\\\n\n= 43.19 \\;mol \\\\\n\nN_A = 6.022 \\times 10^{23} \\\\\n\nn = 43.19 \\times 6.022 \\times 10^{23} \\\\\n\n= 2.6 \\times 10^{25}"
Answer: "2.6 \\times 10^{25}"
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