Question

The system that contains the maximum number of atom is:
(A) $ 4.25{\text{ g of N}}{{\text{H}}_3} $
(B) $ {\text{8 g of }}{{\text{O}}_2} $
(C) $ 2{\text{ g of }}{{\text{H}}_2} $
(D) $ 4\,{\text{g of He}} $

Answer 1

Hint :Atom is the basic unit of matter, also it is the smallest unit of matter that has the characteristic properties of an individual element. In order to calculate the number of atoms in a sample we have to find out how many moles of the element is present in the sample. A mole is just a unit which is equal to Avogadro's number ( $ 6.02 \times {10^{23}} $ ) of atoms .
 $ {\text{No}}{\text{.of atoms = No}}{\text{.of moles }} \times {\text{ N}} \\
  {\text{ N - Avogadro}}{{\text{o}}^,}{\text{s number}}\left( {6.02 \times {{10}^{23}}} \right) \\
  {\text{No}}{\text{.of moles = }}\dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}} \\
  $

Complete Step By Step Answer:
To calculate the number of atoms of a given sample , we have to know the given weight of the sample , its atomic mass and the constant Avogadro's number.
So here first we have to find the number of atoms in $ 4.25{\text{ g of N}}{{\text{H}}_3} $ for finding the No. of atoms first we have to calculate the molar mass then No. of moles . We can find the molar mass of $ {\text{N}}{{\text{H}}_3} $ by
 molar mass of $ {\text{N}}{{\text{H}}_3} $ = $ {\text{Atomic mass of N + 3}} \times {\text{ Atomic mass of H}} $
Atomic mass of nitrogen is $ 14.01\,g $ and that of hydrogen is $ 1.01\,g $ .
So we get
 $
  {\text{Molar mass of N}}{{\text{H}}_3} = \,14.01\, + \,3 \times 1.01 \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14.01\, + \,3.03\, \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 17.04\,g\,{\text{of N}}{{\text{H}}_3} \\
  $
Given mass of $ {\text{N}}{{\text{H}}_3} $ = $ 4.25\,g $
Now we have to calculate No. of moles in $ 4.25{\text{ g of N}}{{\text{H}}_3} $ . It is given by
 $ = \dfrac{{4.25\,g}}{{17.04\,g}} = 0.2494 $ . Here we get the No. of moles , once we get the No. of moles we can calculate the No. of atoms.
So the No. of atoms in $ 4.25{\text{ g of N}}{{\text{H}}_3} $ = $ 0.2494\, \times \left( {6.02 \times {{10}^{23}}} \right) = \,1.5013 \times {10^{23}} $
Here $ 4.25{\text{ g of N}}{{\text{H}}_3} $ contains $ \,1.5013 \times {10^{23}} $ No. of atoms.
Next we have to calculate the No. of atoms in $ {\text{8 g of }}{{\text{O}}_2} $ . We can calculate this also by above method
 $ {\text{Molar mass of }}{{\text{O}}_2}\, = \,\,2 \times 16.00\,g\, = 32\,g $
Where $ 16g $ is the atomic mass of oxygen atom.
So the No. of moles in $ {\text{8 g of }}{{\text{O}}_2} $ = $ \dfrac{{8\,g}}{{32\,g}}\, = \,\,0.25 $
No. of atoms in $ {\text{8 g of }}{{\text{O}}_2} $ = $ \,0.25\, \times \left( {6.02 \times {{10}^{23}}} \right)\, = \,1.505\, \times {10^{23}} $
Here $ {\text{8 g of }}{{\text{O}}_2} $ contains $ \,1.505\, \times {10^{23}} $ No. of atoms.
Another is to calculate No. of atoms in $ 2{\text{ g of }}{{\text{H}}_2} $ . So first we have to calculate the molar mass of $ {{\text{H}}_2} $ .
molar mass of $ {{\text{H}}_2} $ = $ 2 \times 1.01g = 2.02g $
No. of moles in $ 2{\text{ g of }}{{\text{H}}_2} $ = $ \dfrac{{2g}}{{2.02g}}\, = \,0.9900 $
So the No. of atoms in $ 2{\text{ g of }}{{\text{H}}_2} $ = $ 0.9900\, \times \left( {6.02 \times {{10}^{23}}} \right)\, = \,5.959 \times {10^{23}} $
  $ 2{\text{ g of }}{{\text{H}}_2} $ contains $ 5.959 \times {10^{23}} $ No. of atoms.
Next is to find the No. of atoms in $ 4\,{\text{g of He}} $ . Here $ {\text{He}} $ atom is given hence we have to find out the atomic mass from the periodic table.
The atomic mass of $ {\text{He}} $ is $ 4.00g $
So the No. of moles in $ 4\,{\text{g of He}} $ = $ \dfrac{{4g}}{{4.00g}} = \,1 $
Once we get the No. of moles we can calculate the No. of atoms
The No. of atoms in $ 4\,{\text{g of He}} $ = $ 1\, \times \,\left( {6.02\, \times {{10}^{23}}} \right)\, = \,6.02 \times \,{10^{23}} $
So $ 4\,{\text{g of He}} $ contains $ \,6.02 \times \,{10^{23}} $ No. of atoms.
From above calculations it is clear that the system which contains the maximum No. of atoms is $ 4\,{\text{g of He}} $ .
Hence the correct answer is option D.

Note :
Avogadro's number is the number of particles in one mole of any substance. It is equal to $ \,6.02 \times \,{10^{23}} $ . The particles can be electrons , molecules or atoms.
In chemistry the mole is the unit of amount. We can define a mole of a substance as ‘ The mass of a substance containing the same number of fundamental units as there are atoms in exactly $ 12g $ of $ ^{12}{\text{C}} $ . Where the fundamental units are atoms, molecules or formula units.