Question

Which of the following has the maximum number of atoms?
(A) 56 litres of $S{{O}_{2}}$
(B) 44.8 litres of $N{{H}_{3}}$
(C) 33.6 litres of $C{{H}_{4}}$
(D) 56 litres of $CO$

Answer 1

Hint: We can express the mole as the amount of substance. For transforming between moles and grams, we could use the molar mass of the substance and for transforming volume and moles of a gas, we could use the idea of the number of moles existing in 22.4 litre of volume of a gas.

Complete Solution :
   - Let's start with the concept of moles and Avogadro number. Avogadro’s number can be described as a proportion which relates the molar mass on an atomic scale to physical mass on a human scale. It can also be defined as the number of elementary particles such as, compounds, atoms, molecules, etc. per mole of a substance.
  - Or in other words the number of units in one mole of any substance is called Avogadro’s number or Avogadro’s constant and Its value is given as $6.023\times {{10}^{23}}$.

- In terms of volume also the concept of moles can be defined. That is at standard pressure and temperature (STP) conditions, one mole of any gas ($6.023\times {{10}^{23}}$ particles) will occupy a volume of 22.4 litres. Hence, let’s calculate the number of moles in the given options according to the above defined concepts of moles.

- Number of moles in \[56\text{ }litres\text{ }of\text{ }S{{O}_{2}}=\dfrac{56}{22.4} = 2.5\text{ }mole\]
$S{{O}_{2}}$ have three atoms per molecule and hence number of atom in molecule can be written as
 \[Number\text{ }of\text{ }atom\text{ }in\text{ }molecule=3\times 2.5=7.5\text{ }moles\]
Number of moles in \[44.8\text{ }litres\text{ }of\text{ N}{{\text{H}}_{3}}=\dfrac{44.8}{22.4}=2\text{ }mole\]

- $N{{H}_{3}}$ have four atoms per molecule and hence number of atom in molecule can be written as
\[Number\text{ }of\text{ }atom\text{ }in\text{ }molecule = 4\times 2 = 8\text{ }moles\]
- Number of moles in \[33.6\text{ }litres\text{ }of\text{ }C{{H}_{4}} = \dfrac{33.6}{22.4} = 1.5\text{ moles}\]

- $C{{H}_{4}}$ has five atoms per molecule and hence number of atom in molecule can be written as
\[Number\text{ }of\text{ }atom\text{ }in\text{ }molecule = 5\times 1.5 = 7.5\text{ }moles\]

- Number of moles in \[56\text{ }litres\text{ }of\text{ }CO = \frac{56}{22.4} = 2.5\text{ }moles\]
 $CO$ has two atoms per molecule and hence number of atom in molecule can be written as
\[Number\text{ }of\text{ }atom\text{ }in\text{ }molecule=2\times 2.5 = 5\text{ }moles\]
Therefore maximum number of atoms are present in 44.8 litres of $N{{H}_{3}}$
So, the correct answer is “Option B”.

Note: There will be a confusion on calculating the number of moles in volume. If we are given the number of moles then to calculate the volume multiply the number of moles by 22.4 and If we are given the volume of gas in litres or $d{{m}^{3}}$, then to calculate number of moles present we should divide the given volume by 22.4 .