# volume of16 g of oxygen at STP

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Given

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

Formula...

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

At STP

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

=11.2 litres

Therefore 16 gram of oxygen gas has a volume of 11.2 litres at STP