1. A sample of oxygen gas is under a 100.60 kPa pressure, in a 4.69 L container, at 2000C . How many grams of oxygen are present in this sample?

Answer 1

Ideal gas law:

PV = nRT

If we can find n (moles), it will be easy to convert that to grams using the molar mass of oxygen gas.

P = 100.60 kPa

V = 4.69 L

n = ?

R = gas constant = 8.314 L-kPa/K-mole

T = 2000ºC + 273 = 2273K

Solving for n:

n = PV/RT = (100.60 kPa)(4.69 L) / (8.314 L-kPa/Kmol)(2273K)

n = 0.02497 moles of oxygen

Oxygen gas is O₂ and the molar mass = 32 g/mole

Mass of O₂ present = 0.02497 moles O₂ x 32 g/mole = 0.799 g (to 3 significant figures)