From measurements based on the carbon-12 scale the number of atoms in 0.012 kg, or 12.00 g, of C-12 is 6.02 x 1023. From the definition of the mole, one mole of every substance contains this same number of entities. This number is called the Avogadro constant. It has the Symbol NA.
The Avogadro constant has units mol- ( and can be quoted as N A = 6.02 x 1023 entities mol-1.
From the definition of Avogadro’s constant:
1 mol of any substance contains N A entities
½ mol of any substance contains ½ x N A entities
3 mol of any substance contains 3 x N A entities
X mole of any substance contains X x N A entities
. . Number of entities, N, in a given amount of substance, n, is given by:
N = n x NA
n = N / NA
MOLAR MASS (M)
The mass of one mole of particles is called molar mass. The number of entities N A (atoms or molecules) contained in molar mass is equal to 6.023 x 1023. For example,
(i) One mol of sodium means 6.023 x 1023 Na atoms weighing 23.0 g (GAM of sodium). In brief we can write that
Molar mass of sodium = 23.0 g mol-1
(ii) One mol of ammonia contains 6.023 x 1023 NH3 molecules weighing 17.0 g (GMM of ammonia). Thus we can write
Molar mass of ammonia = 17.0 g mol- 1•
6.023 x 1023 atoms of oxygen weigh
= 16 gram = GAM of oxygen or 1g atom
6.023 x 1023 atoms of nitrogen weigh= 14 gram = GAM of nitrogen or 1g atom
6.023 x 1023 molecules of oxygen weigh= 32 gram = GMM of oxygen or 1g molecule
6.0Z3 x 1023 molecules of water weigh= 18 gram= GMM of water or 1g molecule . .
Example 10.6. (i) Calculate the number of atoms in 16 g of Cu?
(ii) How many moles are there in 2.7g of Al?
(Given: atomic mass of Cu = 64u, Al = 27u)
Solution. (i) Number of moles in16 g Cu (u) = 64= Massing g / Atomic mass = 16 / 64 = 1/ 4 = 0.25 mol
Number of atoms = mol x NA = 0.25 x 6.02 x 1923= 1.50 x 1023
(ii) Atomic mass of AI = 27
:. Number of moles in 2.7 g AI (n) = 2.7 / 27 = 1 / 10 = 0.1 mol
Example 10.7. (i) What is the mass of0.600 mol of K2C03?
(ii) What is the amount of substance is 5 g CaCO 3?
(Relative atomic masses are: K = 39, C = 12, Ca = 40).
Solution. (i) Molar mass of k2CO3 = (2 x 39) + 12 + (3 x 16)= 138 g mol-1
Mass of substance = ( amount of / substance ) x ( molar / mass)
Mass of 0.600 mol of k2CO3 = 5 g
(ii) mass of CaCO3 = 5 g
molar mass of CaCO3 = 100 g mol-1
amount of CaCO3 in 5 g = 5 g / 100 g mol-1 = 0.05 mol.
Example 10.7 (i) How many molecules of O 2 gas are present in 16 g of oxygen gas?
(ii) What is the molar mass and relative molecular mass of a substance if the mass of 0.10 mol is 3.9 g?
Solution. (i) mass of O 2 (m) = 16 g
molar mass of O 2 (M) = 32 g mol-1
molar of oxygen , n(O2) = m / M = 16 g / 32 mol -1 = 0.50 mol
Now, No. of molecules (N) =moles (n) x NA
. . number of molecules, N(O2
= 0.50 mol x 6.02 x 1023 molecules mol-1
= 3.01 x 1023 molecules
(ii) mass of substance = 3.9 g
amount of substance = 0.10 mol
molar mass = 3.9 / 0.10 mol
= 39 g mol -1
relative molecular mass = 39.
MOLAR VOLUME (V m)
In 1811, Amedeo Avogadro, an Italian scientist proposed a law known as Avogadro’s Law.
“Avogadro s Law states that equal volumes of all gases contain equal number of molecules under similar conditions.”
In other words, we can say that different samples of gases containing equal number of molecules occupy same volume under similar conditions. As the quantities of chemical entities is expressed in terms of mole, hence it can be deduced that one mole molecules of all gases contain same number (6.023 x 1023 ) of molecules therefore, they occupy same volume under similar conditions of temperature and pressure. The volume occupied by one mole molecules of a gaseous substance is called Molar Volume or Gram Molecular Volume (G.M.V.). One mole molecules of all gases occupy 22.4 litres at 273 K and 760 mm pressure (S.T.P.). Hence, molar volume of all gases at S.T.P. is 22.4 1itres.
Molar volume is expressed by symbol V m and its unit is dm3 moi-1•
The relationship between any given volume of gas V and amount of substance n at standard temperature and pressure (S.T.P) is derived as follows:
1 mol of every gas occupies a volume of 1 x 22.4 dm3
X mol of every gas occupies a volume of X x 22.4 dm3
:. V(dm3) = n x V m
- I. Objective type Questions
Select the most appropriate choice from the options given as
(a), (b), (c) and (d) after each question:
1. Avogadro’s number (NA) represents the number of atoms in
(a) 12 g of C12 (b) 320 g of sulphur
(c) 32 g of oxygen (d) 12.7 g of iodine.
2.The number of moles of carbon dioxide which contain 8 g of oxygen is
(a) 0.50 mol (b) 0.20 mol
(c) 0.40 mol (d) 0.25 mol.
3.The number of atoms present in 0.5 g-atom of nitrogen is same as the atoms in
(a) 12 g of carbon (b) 32 g of sulphur
(c) 8 g of the oxygen (d) 24 g of magnesium.
4.An atom is 10 times heavier than l/12th of mass of a carbon atom (C-12). The mass of the atom in u is
(a) 10 (b) 120
(c) 1.2 (d) 12.
5. The total number of ions present in 111 g of CaCl2 is
(a) one mole (b) two mole
(c) three mole (d) four mole.
6. Which of the following weighs the most?
(a) One g-atom of nitrogen
(b) One mole of water
(c) One mole of sodium
(d) One molecule of H2S04.