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Chemistry

Molar Volume

Molar volume is the volume occupied by one mole of any gas at a definite pressure and temperature. It is denoted by Vm. Molar volume of the substance depends on temperature and pressure. The unit of molar volume is litre per mol or millilitre per mol.

As per Avogadro’s law, equal volumes of all gases contain equal number of molecules, at a constant temperature and pressure. Therefore, equal number of molecules of any gas, must occupy the same volume, at constant temperature and pressure.

Standard Molar Volume

Standard molar volume of a gas is the volume occupied by 1 mole of any gas at 273 K and 1 atm pressure (STP). It is equal to 22.4 litres of 22,400 ml. It is the same for all gases.

Remember

S.T.P. = Standard Temperature and Pressure

Standard Temperature = 0oC or 273 K

Standard Pressure = 1 atm or 760 mm of mercury

Calculation of Molar Volume

Example of oxygen

Mass of 1 litre of oxygen at STP = 1.429 g

Mass of 1 mol of oxygen = 32 g

Volume : Mass

1 litre : 1. 429 g

x : 32 g

x= 32/1.429

x= 22.4 Litres

The following table gives the relation between the Gram Molecular Weight (GMW), Number of moles, Molar Volume and the Number of particles for gases at STP.

Relationship between Various Parameters of Gases, at STP

GasMolecular FormulaGMW
(in g)
No.Of MolesMolar Volume
dm3 or l
No.of moles in 1 mole
HydrogenH22122.46.023x 1023
OxygenO232122.46.023x 1023
NitrogenN228122.46.023x 1023
ChlorineCl271122.46.023x 1023
Carbon dioxideCO244122.46.023x 1023
Nitrogen dioxideNO246122.46.023x 1023
AmmoniaNH317122.46.023x 1023
MethaneCH416122.46.023x 1023
Sulphur dioxideSO264122.46.023x 1023

Problems Based on Molar Volume

Example:

Calculate the volume occupied by 3.4 g of ammonia at STP. (N=14, H=1)

Solution

Gram molecular mass of ammonia (NH3) = (1 x 14) + (3 + 1) = 14 + 3 = 17 g

Mass of 1 mol of ammonia = 17g

Molar volume = 22.4 litres

Volume of 3.4 g of ammonia at STP = ?

Mass : Volume

17 g : 22.4 litre

3.4 g : x

x= (3.4 X 22.4)/17

x= 4.48 Litres

Volume occupied by 3.4 g of ammonia at STP =4.48 litres

Example:

56 ml of carbon dioxide has a mass at 0.11 g at STP. What is the molar mass of the carbon dioxide?

Solution:

Mass of 56 ml of carbon dioxide at STP = 0.11 g

Mass of 22400 ml of carbon dioxide = ?

Mass : Volume

0.11 g : 56 ml

x g : 22400 ml

x = (0.11 x 22400)/56

x = 44g

Mass of 22400 ml of carbon dioxide at STP = 44 g

Molar Mass of carbon dioxide = 44 g/mole.

Example:

One gram of pure sulphur dioxide has a volume of 350 ml at STP. What is the Relative Molecular Mass of sulphur dioxide?

Solution:

Volume of sulphur dioxide gas = 350 ml at STP

Mass of sulphur dioxide gas = 1 g

Mass of one mole of sulphur dioxide = x g/mole

Volume of 1 mole of sulphur dioxide = 22400 ml at STP

Mass : Volume

1 : 350 ml

x : 22400 ml

x= (22400 x 1)/350 = 64

Mass of 1 mole of SO2 = 64 g/mole

Relative molecular mass of sulphur dioxide = 64.

Deduction of relationship between molecular mass and vapour density

Relative Molecular Mass

Relative molecular mass is the ratio of the mass of one molecule of a substance to the mass 1/12th

of a carbon atom, or 1 amu.

Vapour Density

Vapour density is the ratio of the mass of a volume of a gas, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.

vapour density = mass of n molecules of gas / mass of n molecules of hydrogen. (and thus: molar mass = ~2 × vapour density) For example, vapour density of mixture of NO2 and N2O4 is 38.3 . Vapour density is a unitless quantity.

Avogadro’s Law

According to Avogadro’s law, equal volumes of all gases contain equal number of molecules.

Thus, let the number of molecules in one volume = n,

vapour density

or,

Relative Molecular Mass = 2 x Vapour density

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