Question

How many grams of O₂ would be in a 4.6 L container at 510.1 mm Hg at 37.3^o C?

Answer 1

We know the relationship from gas law

PV = nRT -------(1)

P = Pressure in atm, V = volume in Litre , n = number of moles

R = gas constant = 0.082 atm L /mol.K , T is Temperure in Kelvin

P = 510.1 mm Hg = 510.1 atm / 760 = 0.671 atm

V = 4.6 L,

T = 37.3 oC = (37.3 + 273.15)K = 310.45 K

From (1)

0.671 atm * 4.6 L = n * 0.082 atm L mol-1K-1 * 310.45 K

3.0866 = n * 25.457 mol^-1

n = 3.0866 / 25.457 mol^-1 = 0.1212 mol

Molar mass of O2 = 32 g/mol

Mass of O2 = 32 g/mol * 0.1212 mol =3.88 g

Mass of O2 in gram = 3.88 g