Mole Concept (Avogadro Constant) , Molar mass, Percentage Composition,Emprical and Molecular Force

Solved Example

Question 1 Calculate the no. of moles present in 6 gm Carbon.
Solution
For Carbon
G. A. M. = 1 mole
12 gm = 1 mole
1 gm = 1/12mole
For 6 gm = 6 x (1/12)moles
=1/2 moles.

Question 2 Calculate the no. of moles in 2.8 gm N?
Solution
G. A. M. = 1 mole
14 gm = 1 mole
1 gm = 1/14 moles
2.8 gm = 2.8 *(1/14) moles
= 0.2 moles

Question 3Give the no. of atoms present in five moles N
Solution
Moles N_a = 6.022 × 10²³ atoms
5 moles N_a =30.11 × 10²³ atoms

Question 4Calculate the molesin 10 gm Carbon atom
Solution
1Mole = G. A. M.
1Mole = 12 gm
1 gm = 1/12 mole
10 gm= 10* (1/12) mole
=10/12 mole

Question 5Find the mass occupied by 0.5 mole N.
Solution
1 mole = Gram atomic mass
1 mole = 14 gm
0.5 mole= 14 *0.5
= 7 gm

Question 6 Give the no. of atoms present in 2.5 moles carbon
Solution
1 mole =N_A atoms
1 mole = 6.022 × 10²³ atoms
So, 2.5 moles = 6.022 *2.5 * 10²³
= 15.0490 × 10²³ atoms
= 15.049 × 10²³ atoms

Question 7Give the no. of moles present in 3.01 × 10²³ atoms of Sodium
Solution
1 mole =N_A atoms
1 mole = 6.022 × 10²³ atoms
1 atom = ( 1/ 6.022 × 10²³ ) moles
So, in 3.01 × 10²³ atoms =3.01× 10²³ × ( 1/ 6.022 × 10²³ ) = .5 Mole

Question 8Give the no. of moles present in 0.6 gm Carbon.
Solution
G. A. M. = 1 mole
12 gm = 1 mole
1 gm =1/12 mole
So, in 0.6 gm = .6 X (1/12) mole
= 0.05 mole

Question 9 Calculate the no. of atoms present in 2.3 gm Sodium.
Solution
N_A atoms = G. A. M.
6.022* 10²³ atoms = 23 gm
23 gm= 6.022* 10²³ atoms
1 gm = (1/23) * 6.022* 10²³ atoms
2.3 gm = (2.3/23) * 6.022* 10²³ atoms
=6.022* 10²²atoms

Question 10 Give the G. A. M. of an element occupying 1.21 volume at STP is 3.4 gm.
Solution
For 11.2 L = 3.4 gm
Multiplying it by ‘2’
22.4 L = 6.8 gm
So G.A.M =6.8 gm

Question 11 Give the no. of Molecules present in 0.18 gm of water.
Solution
Water Molecular mass =18
so 18 gm = 1moles
0.18 g = .18 * (1/18) moles
= 0.01 mole
=6.022 * 10²¹ molcules

Example
C₆ H₁₂ O₆ =(Glucose)
% of C is given by

=$ \frac {6 * 12}{180} * 100$
= 40%
% of Hydrogen is
=$ \frac {12 *1}{180} * 100$
=6.67%
% of Oxygen
=$ \frac {6* 16}{180} * 100$
=53.33%