Sunday, 4 January 2015

3.4 Number of Moles and Its Molar Volume

Number of Moles and Its Molar Volume
Definition : : Molar volume : The volume occupied by 1 mole of gas at a particular temperature and pressure.

At room conditions (25°C and pressure 1 atmosphere), 1 mole of gas occupies a volume of 24dm3 .

At standard condition (0 and pressure 1 atmosphere), 1 mole of gas occupies a volume 22.4dm3.
■ Summary of relationship between the number of moles, the volume of gas and the molar volume of gas.

Worked-example 3.4(a)

Calculate the volume occupied by 3.5 moles of ammonia gas, NH3 at s.t.p.
Solution

Volume of gas
= 3.5×22.4
= 78.4dm3

Worked-example 3.4(b)

Calculate the volume occupied by the following gases at s.t.p:
[RAM: H = 1, C = 12, N = 14, O = 16,], [NA = 6.02×1023 , 1 mole gas occupies 22.4dm³ at s.t.p]

(a) 8.8 g carbon dioxide, CO2 
Solution
Step 1:
Number of moles = Mass of Carbon Dioxide Relative molecular mass
=8.8 12 + (2×16)
=8.8 12 + 32
=8.8 44
= 0.2mol
Step 2:
Volume of 8.8 g carbon dioxide, CO2
= Number of mole×Molar volume
= 0.2×22.4
= 4.48dm3

(b) 1.5×1023 molecules of oxygen, O2 
Solution
Step 1:
Number of moles = Number of particles Avogrado constant
=1.5×1023 6×1023
= 0.25mol
Step 2
Volume of oxygen
= Number of mole×Molar volume
= 0.25×22.4
= 5.6dm3
Summary of Mole Concept
■ Summary of relationship between the number of moles of compound with the number of particles, particle mass and gas volume.

Worked-example 3.4(c)

Calculate the volume occupied by the following gases at s.t.p.
[Molar volume at s.t.p = 22.4dm3, NA = 6×1023 , H = 1, C = 12, O = 16, N = 14, Cl = 35.5, Ne = 20]

(a) 0.5 g of neon , Ne.
Solution
Step 1:
Number of moles = Mass of neon Relative atom mass
=0.5 20
= 0.025mol
Step 2:
Volume of 0.5 g neon CO2
= Number of mole×Molar volume
= 0.025×22.4
= 0.56dm3

(b) 1.5 moles of carbon dioxide, CO2
Solution :
Volume of 1.5 moles of CO2
= Number of mole×Molar volume
= 1.5×22.4
= 33.6dm3

(c) 4.8×1022 molecules Chloride, Cl2
Solution
Step 1:
Number of moles = Number of particles Avogrado constant
=4.8×1022 6×1023
= 0.08mol
Step 2:
Volume of Cl2
= Number of mole×Molar volume
= 0.08×22.4
= 1.792dm3


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