Question

Which one of the following has the maximum number of atoms of oxygen?
A. ${\rm{2}}\;{\rm{g}}\;{\rm{of}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}$
B. ${\rm{2}}\;{\rm{g}}\;{\rm{of}}\;{\rm{S}}{{\rm{O}}_2}$
C. ${\rm{2}}\;{\rm{g}}\;{\rm{of}}\;{\rm{C}}{{\rm{O}}_2}$
D. ${\rm{2}}\;{\rm{g}}\;{\rm{of}}\;{\rm{CO}}$

Answer 1

We know that the number of atoms can be determined by using two quantities one of them is Avogadro’s number and the other one is the respective molar mass of all the species.

Complete step by step solution:
The general formula for determining the number of atoms by using the formula is shown below.
${\rm{Number}}\;{\rm{of}}\;{\rm{atoms}} = \dfrac{{{{\rm{N}}_{\rm{A}}}}}{{{\rm{Molar}}\;{\rm{mass}}}}.......\left( {\rm{I}} \right)$
Where, ${{\rm{N}}_{\rm{A}}}$ is the Avogadro’s constant and the value of it is $6.023 \times {10^{23}}$.
So, the molar mass of water is ${\rm{18}}\;{\rm{g/mol}}$.
The number of atoms of oxygen present in ${\rm{2}}\;{\rm{g}}\;{\rm{of}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}$ can be calculated by substituting all the respective values in equation (I).

$${\rm{Number}}\;{\rm{of}}\;{\rm{atoms}}\;{\rm{for}}\;{\rm{1}}\;{\rm{g}}\;{\rm{of}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}} = \dfrac{{6.023 \times {{10}^{23}}}}{{{\rm{18}}\;{\rm{g/mol}}}}\\\ = 0.335 \times {10^{23}}\\\ {\rm{Number}}\;{\rm{of}}\;{\rm{atoms}}\;{\rm{for}}\;2\;{\rm{g}}\;{\rm{of}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}} = 0.335 \times {10^{23}} \times 2\\\ = 0.67 \times {10^{23}}$$

So, the molar mass of ${\rm{S}}{{\rm{O}}_{\rm{2}}}$ is ${\rm{64}}\;{\rm{g/mol}}$.
The number of atoms of oxygen present in ${\rm{2}}\;{\rm{g}}\;{\rm{of}}\;{\rm{S}}{{\rm{O}}_2}$ can be calculated by substituting all the respective values in equation (I).

$${\rm{Number}}\;{\rm{of}}\;{\rm{atoms}}\;{\rm{for}}\;{\rm{1}}\;{\rm{g}}\;{\rm{of}}\;{\rm{S}}{{\rm{O}}_2} = \dfrac{{6.023 \times {{10}^{23}}}}{{{\rm{64}}\;{\rm{g/mol}}}}\\\ = 0.094 \times {10^{23}}\\\ {\rm{Number}}\;{\rm{of}}\;{\rm{atoms}}\;{\rm{for}}\;2\;{\rm{g}}\;{\rm{of}}\;{\rm{S}}{{\rm{O}}_2} = 0.094 \times {10^{23}} \times 2\\\ = 0.188 \times {10^{23}}$$

So, the molar mass of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ is ${\rm{44}}\;{\rm{g/mol}}$.
The number of atoms of oxygen present in ${\rm{2}}\;{\rm{g}}\;{\rm{of}}\;{\rm{C}}{{\rm{O}}_2}$ can be calculated by substituting all the respective values in equation (I).

$${\rm{Number}}\;{\rm{of}}\;{\rm{atoms}}\;{\rm{for}}\;{\rm{1}}\;{\rm{g}}\;{\rm{of}}\;{\rm{C}}{{\rm{O}}_2} = \dfrac{{6.023 \times {{10}^{23}}}}{{{\rm{44}}\;{\rm{g/mol}}}}\\\ = 0.136 \times {10^{23}}\\\ {\rm{Number}}\;{\rm{of}}\;{\rm{atoms}}\;{\rm{for}}\;2\;{\rm{g}}\;{\rm{of}}\;{\rm{C}}{{\rm{O}}_2} = 0.136 \times {10^{23}} \times 2\\\ = 0.272 \times {10^{23}}$$

So, the molar mass of ${\rm{CO}}$ is $28\;{\rm{g/mol}}$.
The number of atoms of oxygen present in ${\rm{2}}\;{\rm{g}}\;{\rm{of}}\;{\rm{CO}}$ can be calculated by substituting all the respective values in equation (I).

$${\rm{Number}}\;{\rm{of}}\;{\rm{atoms}}\;{\rm{for}}\;{\rm{1}}\;{\rm{g}}\;{\rm{of}}\;{\rm{CO}} = \dfrac{{6.023 \times {{10}^{23}}}}{{28\;{\rm{g/mol}}}}\\\ = 0.215 \times {10^{23}}\\\ {\rm{Number}}\;{\rm{of}}\;{\rm{atoms}}\;{\rm{for}}\;2\;{\rm{g}}\;{\rm{of}}\;{\rm{CO}} = 0.215 \times {10^{23}} \times 2\\\ = 0.43 \times {10^{23}}$$

Thus, from the calculated values of all the option we get the maximum number of oxygen present in ${\rm{2g}}\;{\rm{of}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}$.
Therefore, the correct option for this question is A that is ${\rm{2g}}\;{\rm{of}}\;{{\rm{H}}_{\rm{2}}}{\rm{O}}$

Note:
The Avogadro constant was proposed by renowned Italian scientist named Amedeo Avogadro in $1811$. It is used to identify the no. of molecules or atoms of the species present in a given mass of substance .