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Question

Question: Which of the following has the largest number of atoms? A.\(4g{\text{ of }}{{\text{H}}_2}\) B.\(...

Which of the following has the largest number of atoms?
A.4g of H24g{\text{ of }}{{\text{H}}_2}
B.16g of O216g{\text{ of }}{{\text{O}}_2}
C.28g of N228g{\text{ of }}{{\text{N}}_2}
D.18g of H2O18g{\text{ of }}{{\text{H}}_2}O

Explanation

Solution

To solve this question, we must use the mole concept. First, we calculate the number of moles and then we can apply the mole concept. We must also know the concept of molecular mass to solve this question which is the sum of the atomic masses of all its constituent elements. Remember that the value of Avogadro number is given 6.02×10236.02 \times {10^{23}} .

Complete step-by-step answer: We know, one mole of a substance contains 6.02×10236.02 \times {10^{23}}atoms.
We also know that, number of moles =m MM    = \dfrac{m}{\ MM \\\ \\\ \ } where, m is the given mass and MM represents molar mass.
Therefore, first we need to calculate the number of moles present in 4g of H24g{\text{ of }}{{\text{H}}_2}.
Hence, the number of moles of H2=42=2{H_2} = \dfrac{4}{2} = 2moles of H2{H_2}. Since 11 mole of H2{H_2} =6.02×1023 = 6.02 \times {10^{23}} H2{H_2} atoms.
Therefore, 22 moles of H2{H_2} will contain 2×6.02×1023=12.04×10232 \times 6.02 \times {10^{23}} = 12.04 \times {10^{23}}atoms.
Similarly, number of moles in 16g of O216g{\text{ }}of{\text{ }}{O_2} =1632=0.5 = \dfrac{{16}}{{32}} = 0.5moles.
0.50.5 moles of O2{O_2} will contain = 0.5×6.02×10230.5 \times 6.02 \times {10^{23}} =3.01×1023 = 3.01 \times {10^{23}}atoms
Now, number of moles in 28 g of N2=28{\text{ }}g{\text{ }}of{\text{ }}{N_2} = 2828=1\dfrac{{28}}{{28}} = 1mole of N2. We know that 11 mole of N2{N_2} contains 6.02×10236.02 \times {10^{23}}atoms.
Lastly, number of moles in18 g of H2O = 118{\text{ }}g{\text{ }}of{\text{ }}{H_2}O{\text{ }} = {\text{ }}1. Hence, the number of atoms in 18 g of H2O 18{\text{ }}g{\text{ }}of{\text{ }}{H_2}O{\text{ }} =6.02×1023 = 6.02 \times {10^{23}}atoms of H2O.
Therefore, we can conclude that the largest number of atoms is present in 4g of H24g{\text{ of }}{{\text{H}}_2} . It contains 12.04×102312.04 \times {10^{^{23}}}atoms. Rest all options contain less number of atoms.

Hence, option A is the correct option.

Note: You should be aware about the relationship between no. of moles, molar mass, and atoms of any given substance in order to solve this question. The relationship can be summarized as-
Molar mass of any substance is one mole of that substance which is the Avogadro constant =6.02×1023 = 6.02 \times {10^{23}}atoms or ions or molecules.