Question

"How many molecules are in 1.0 cm3 of air at STP? (R = 8.31 J/mol . K, N A = 6.022 × 1023 molecules/mol)"

Answer 1

Answer:

2.7099×10^19 molecules

Explanation:

At STP,

P = 1 atm = 101325N/m^2, T = 273K

From the ideal gas equation

PV = nRT

n = PV/RT

V = 1cm^3 = 1cm^3 × 1m^3/10^6cm^3 = 10^-6m^3, R = 8.31J/molK

n = 101325×10^-6/8.31×273 = 4.5×10^-5 mole

From Avogadro's law

1 mole = 6.022×10^23 molecules

4.5×10^-5 mole = 4.5×10^-5 × 6.022×10^23 = 2.7099×10^19 molecules