Question

Which of the following contain the maximum number of molecules?
A. $1g{\text{ }}C{O_2}$
B. $1g{\text{ }}{N_2}$
C. $1g{\text{ }}{H_2}$
D. $1g{\text{ }}C{H_4}$

Answer 1

We can solve the above problem with the help of mole concept. According to it, the mass of one mole of atoms is equal to the gram atomic mass. The formula used is:
Number of molecules in “w” gram of an element =$\dfrac{{w \times 6.023 \times {{10}^{23}}}}{{gram{\text{ }}molecular{\text{ }}mass}}$

Complete step by step answer: In case of $1g$ of $C{O_2}$, let us calculate the gram molecular mass of $C{O_2}.$ As we know the gram atomic mass of $C$ is $12g$ and the gram atomic mass of $O$ is $16g.$ Then the gram atomic mass of $C{O_2}$ is calculated as given below:
$12 + 2 \times 16 \\\ \Rightarrow 12 + 32 = 44g \\\$
Number of molecules in $1g$ of $C{O_2}$ is calculated as below:
$\dfrac{{1 \times 6.023 \times {{10}^{23}}}}{{44}}$
$\Rightarrow \dfrac{{6.023 \times {{10}^{23}}}}{{44}} = 0.1368 \times {10^{23}}$
Now in the case of $1 g$ of $N_2$, as we know the gram atomic mass of $N_2$ is $28g.$ Then the number of molecule is calculate as below:
$\dfrac{{1 \times 6.023 \times {{10}^{23}}}}{{28}} = 0.21 \times {10^{23}}$
For the case of 1g of $H_2$ ,the gram atomic mass of $H_2$ is $2g.$ The the number of molecules is calculated as below:
$\dfrac{{1 \times 6.023 \times {{10}^{23}}}}{2} = 3.011 \times {10^{23}}$
For 1g $C{H_4}.$The gram atomic mass of methane is
$12 + 4 = 16g$
Number of molecules are:
$\dfrac{{1 \times 6.023 \times {{10}^{23}}}}{{16}} = 0.376$
Hence the correct answer is option C.

Note: The number of molecules can be alternatively calculated as:
If we take the case of $C{O_2},$ $44g$ of the $C{O_2}$ contains molecules=$6.023 \times {10^{23}}$. So, the number of molecules in $1 g$ of carbon dioxide is calculated as below:
$\dfrac{{6.023 \times {{10}^{23}}}}{{44}} = 0.136 \times {10^{23}}$
Now, if we consider the case of $N_2,$ $28g$ of nitrogen contains molecules =$6.023 \times {10^{23}}$. So $1 g$ of nitrogen contains
$\dfrac{{6.023 \times {{10}^{23}}}}{{28}} = 0.215{\text{ molecules}}$
For the case of $H_2$, 2g of hydrogen contains molecules= $6.023 \times {10^{23}}$. So 1 g of hydrogen contains
$\dfrac{{6.023 \times {{10}^{23}}}}{2} = 3.0115{\text{ molecules}}$
For the case of $CH_4$, 16g of the methane contain molecules=$6.023 \times {10^{23}}$. So $1 g$ of methane contains
$\dfrac{{6.023 \times {{10}^{23}}}}{{16}} = 0.376{\text{ molecules}}$