Examples based on
Avogadro’s Law
EXAMPLE 1.
100 mL of a gas ‘A’ contains x molecules. How many molecules of
the gas ‘B’ will be present in 50 mL of ‘B’ under same conditions of
temperature and pressure?
Solution: A : B
Volume: 100 mL 50 mL
No. of molecules: x
?
According to the Avogadro’s law, equal volumes of two gases
under similar conditions would contain the same number of molecules. Therefore,
100 mL of gas B would contain x molecules.
Hence, 50 mL of gas B would contain = x × 50 / 100 molecules = x
/ 2 molecules
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EXAMPLE 2.
Samples of the gases, oxygen, nitrogen, carbon dioxide and
carbon monoxide under the same conditions of temperature and pressure contain
the same number of molecules, (say X).
The molecules of oxygen (O2) occupy V litres and have a mass of 8 g.
(a) What is the volume occupied by
i.
X
molecules of nitrogen (N2)?
ii.
2X molecules of carbon monoxide (CO)?
(b) What is the mass of CO2 in grams?
(Atomic masses
are : C = 12, N = 14, O = 16)
Solution: The given data are rearranged as follows:
Gas: Oxygen Nitrogen
Carbon Carbon
dioxide monoxide
No. of molecules: X X X X
Volume: V
L
Mass of gas: 8 g
Equal volumes of gases under similar conditions of temperature
and pressure contain the same number of molecules. Therefore,
(a) (i)
Volume occupied by X molecules of N2
= V L
(ii)
Volume occupied by 2X molecules of CO
= 2 × V L = 2V L
(b) On molar basis; O2 º CO2
1 mol 1
mol
2 × 16 g = 32 g (12 g + 2 × 16 g) = 44 g
Therefore, 32 g
of O2 º 44 g of
CO2
So, 8 g
of O2 = (44 g / 32 g) × 8 g of CO2 = 11 g
 |
EXAMPLE 3.
The volumes of gases A, B, C and D are in the ratio of 1 : 2 : 2
: 4
under the same conditions of temperature and pressure.
a. Which sample of gas contains the maximum number of
molecules?
b. If the temperature and pressure of gas A are kept
constant, then what will happen to the volume of A, when the number of its
molecules is doubled?
c. If the volume of A is actually 5.6 L at STP,
calculate the number of molecules in the actual volume of D at STP. [Avogadro’s
number: 6.02 × 1023]
d. Using your answer in (d), state the mass of D, if the
gas is dinitrogen oxide (N2O). (N = 14, O =16).
Solution: The given data are written as follows:
Gas: A B
C D
Volume ratio: 1 :
2 : 2 : 4
a. The number of molecules is proportional to the
volume.
Hence, gas D will contain the maximum number of
molecules.
b. Volume of A will get doubled since the number of
molecules depends directly on the volume.
c. Volume of A at STP = 5.6 L
So, Volume of D at STP = 4 × 5.6 L = 22.4 L
22.4 L of any gas at STP will contain one Avogadro’s
number of molecules. So, No. of molecules in the actual volume of D = 6.02 × 1023
d. The mass of any gas occupying a volume of 22.4 at STP
is equal to its molar mass. Hence
Mass of gas D = Mass of 1 mole of N2O
= (2 × 14 +
1 × 16) g = (28 + 16) g = 44 g
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EXAMPLE 4.
The gases chlorine, nitrogen, ammonia and sulphur dioxide are
collected under the same conditions of temperature and pressure. The following
table gives volume of gases collected and the number of molecules x in 20 L of nitrogen. Complete the
table giving the number of molecules in the other gases in terms of x.
Gas Volume/L No. of molecules
Chlorine 10 —
Nitrogen 20 x
Ammonia 20 —
Sulphur dioxide 5 —
Solution: The number of molecules in any sample of a gas is
proportional to its volume.
So,
Gas Volume/L No. of molecules
Chlorine 10 x / 2
Nitrogen 20 x
Ammonia 20 x
Sulphur dioxide
5 x / 4
