volume of16 g of oxygen at STP
Mass of Oxygen = 16 gm
And STP condition that is ....
P (Pressure) = 101.3 kpa
R (gas constant) = 8.314 Lkpa/k mol
T (absolute temperature) = 27. k
V (volume) = ?
n (mole) = mass / molar mass = 16/32 = 0.5 mol
PV = nRT
101.3 * V = 0.5 * 8.314 * 273
101.3 * V = 1134.861
V = 1134.861/101.3
V = 11.20 Lts.
11.20 lts is the amount of volume occupied by 16 gm of Oxygen at STP condition.
at STP, the volume of 1 mole any gas(that is 1 mole of any gas will occupy) 22.4 L.
Moles = mass/molar mass
mass of oxygen gas (O2) = 16 g and molar mass of oxygen is 32 g/mol
Moles of oxygen gas = 16 g/(32 g/mol) = 0.5 mol O2
So the volume of 0.5 mol O2 would be
0.5 mol O2 * (22.4 L / 1 mol O2)
11.2 L o2.
So the volume of 16 g of O2 in STP would be 11.2 L
We have from Avogadro's Hypothesis
1 mole of any gas has a volume of 22.4 litres
thus,1 mole of oxygen gas has a volume of 22.4 litres
thus,16*2 gram of oxygen gas has a volume of 22.4 litres
thus,1 gram of oxygen gas has a volume of 22.4/16/2 litres
thus,16 gram of oxygen gas has a volume of 22.4/16/2*2 litres
Therefore 16 gram of oxygen gas has a volume of 11.2 litres at STP