Sunday 31 January 2021

Counting Atoms-Molar Volume (IGCSE CHEMISTRY)

 

CH. 9 COUNTING ATOM
***MOLAR VOLUME**


KKCS International Program

Chemistry IGCSE 1

Mr. Markus


MOLAR VOLUME


  • The volume of one mole of gas is called the molar volume


  • At r.t.p. [room temperature (25°C) and pressure (1 atmosphere)] :

one mole of any gas occupies a volume of 24 dm3, at r.t.p. the molar volume of a gas is 24 dm3


The figure below shows the molar volumes of three gases at r.t.p.

 

1 mol H2 1 mol NH3 1 mol CO2

(2.0 g) (17.0 g)   (44.0 g)

6x1023 6x1023 6x1023    2 molecules       NH3 molecules      CO2 molecules

24 dm3


24 dm3


24 dm3 At s.t.p. [standard temperature (0°C) and pressure (1 atmosphere)] :

one mole of any gas occupies a volume of 22.4 dm3, at s.t.p. the molar volume of a gas is 22.4 dm3




  • Number of moles of a gas at r.t.p. can be determined from the following formula:





  • The volume occupies by a certain number of moles of a gas at r.t.p can be  found from the following formula:


Example 1


  • Calculate the number of moles of gas particles in each of the following gases at r.t.p

(a) 200 cm3 of oxygen gas

(b) 10 dm3 of ammonia gas

Solutions:

  1. 200 cm3 = 0.2 dm3

number of moles of O2 molecules

= volume/molar volume

=        0.2 dm3 = 0.0083 mol

    (24 dm3 mol -1)

(b)  Number of moles of NH3 molecules

= volume/molar volume

=        10 dm3 = 0.42 mol

    (24 dm3 mol -1)


Example 2


  • Find the volume occupies by the following gases at  r.t.p :

(a) 0.05 mol of methane

(b) 4.0 mol od propane

Solutions:

  1. Volume of methane

= (moles) x (molar volume)

= 0.05 mol x 24 dm3/mol

= 1.2 dm3

  1. Volume of propane

= (moles) x (molar volume)

= 4.0 mol x 24 dm3/mol

= 96 dm3

Example 3


  • Calculate the volume occupied by the following gases at s.t.p :

(a) 0.030 of carbon monoxide

(b) 1.40 mol of nitrogen monoxide

Solutions:

  1. Volume of carbon monoxide

= (moles) x (molar volume)

= 0.030 mol x 22.4 dm3/mol

= 0.672 dm3

  1. Volume of nitrogen monoxide

= (moles) x (molar volume)

= 1.40 mol x 22.4 dm3/mol

= 31.4 dm3


Example 4


  • Calculate the number of moles of gas molecules in each of the following gases at s.t.p

(a) 503 cm3 of fluorine

(b) 12 dm3 of chlorine

Solutions:

  1. 503 cm3 = 0.503 dm3

number of moles of F2 molecules

= volume/molar volume

=        0.503 dm3 = 0.0225 mol

    (22.4 dm3 mol -1)

(b)  Number of moles of Cl2 molecules

= volume/molar volume

=        12 dm3 = 0.54 mol

    (22.4 dm3 mol -1)


Practice


  1. Calculate the volume occupied by 1.0 g  of hydrogen gas at r.tp 

  2. What is the mass of 480 cm3 of nitrogen dioxide gas at r.t.p


Avogadro’s Law


“ At the same temperature and pressure, equal volumes of all gases contain the same number of particles” 


Example:

At room temperature and pressure,

24 dm3 of neon gas contains 1 mole of Ne atoms, i.e. 6 x 10 23 gas particles.

24 dm3 of hydrogen gas contains 1 mole of H2 molecules, i.e. 6 x 10 23 gas particles.

24 dm3 of carbon dioxide gas contains 1 mole of CO2 molecules, i.e. 6 x 10 23 gas particles.


Example 1


  • What is the volume of oxygen needed for the complete combustion of 200 cm3 of hydrogen? 

[all volumes are measured at the same temperature and pressure]

solution;:

Equation of reaction:

2H2(g) + O2(g)  ­čí¬ 2H2O(l)

2 moles of hydrogen need 1 mole of oxygen for complete combustion.


1 mole of hydrogen needs ½ moles of oxygen for complete combustion

At the same temperature and pressure

1 volume of hydrogen needs ½ volume of oxygen

So, 200 cm3 of hydrogen need ½ (200) = 100 cm3 of oxygen


Example 2


  • Benzene is an additive of non-leaded petrol. It has molecular formula of C6H6. What is the volume of oxygen needed for the complete combustion of 10 cm3 of benzene vapour? (all gas volume are measured at the same temperature and pressure)

Solution:

2C6H6(g) + 15O2(g) ­čí¬ 12 CO2(g) + 6H2O(l)

2 moles of benzene need 15 moles of oxygen for complete combustion

1 mole of benzene needs 15/2 moles of oxygen for complete combustion

At the same temperature and pressure

1 vovume of benzene needs 7.5 volumes of oxygen

So, 10 cm3 of benzene need 10 (7.5) = 75 cm3 of O2


Practices


  1. Calculate the volume occupied by each of the following mixture of gases at r.t.p:

(a) 4.0 g of methane and 7.1 g of chlorine

(b) 0.9 mole of nitrogen and 2.1 moles of hydrogen

  1. Find the ratio of the number of molecules in 4.0 g of hydrogen to the number of molecules in 6.0 dm3 of oxygen at r.t.p

  2. At r.t.p., 600 cm3 of oxygen contain n molecules. State, in terms of n the number of

(a) atoms in 16.0 g of sulphur dioxide



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