Question

Number of atoms in \[4.25\] g of ${\text{N}}{{\text{H}}_{\text{3}}}$ is:
A.$6.023 \times {10^{23}}$
B.$4 \times 6.023 \times {10^{23}}$
C.$1.7 \times {10^{24}}$
D.$4.5 \times 6.023 \times {10^{23}}$

Answer 1

Hint: One mole of a chemical entity can be defined as equal to $6.023 \times {10^{23}}$ particles of that substance, where “particles” can mean “atoms, molecules, and the different subatomic particles”. We shall calculate the moles of the compound given and thus, the number of particles.

Formula Used: ${\text{moles}} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$

Complete step by step answer:
The molecular weight of ammonia is equal to the sum of the atomic weights of nitrogen and hydrogen atoms = $\left[ {14 + \left( {3 \times 1} \right)} \right]$= 17 grams
Hence, 17 grams of ammonia = $6.023 \times {10^{23}}$ molecules of ammonia.
Therefore, \[4.25\]g of ammonia = $\dfrac{{6.023 \times {{10}^{23}}}}{{17}} \times 4.25$= $1.505 \times {10^{23}}$ molecules of ammonia.
Now each molecule of ammonia has four atoms, one atom of nitrogen and three atoms of hydrogen. Hence, the total number of atoms in $1.505 \times {10^{23}}$ molecules of ammonia = $4 \times 1.505 \times {10^{23}}$= $6.023 \times {10^{23}}$ atoms.

Hence, the correct answer is option A.

Note:
- The gram atomic weight of an element and the gram molecular weight of a compound is equal to the Avogadro’s number which is equal to $6.023 \times {10^{23}}$.
- The atomic mass of an element is defined as the mass of one atom of the element while the molecular mass of an element is defined as the mass of one molecule of a compound.
- In case of gases the molar mass of the gases is equal to the $22.4$ litre of that gas at standard conditions of temperature and pressure. The vapour density of a gas is equal to the molecular weight of the gas divided by 2. The concept of mole was introduced due to the fact that the mass of the atom is so small that it cannot be weighed and hence a reference was needed. So, one mole of the substance is taken as that reference.