
Which of the following has the largest number of oxygen atoms?
a) $3.65 \times {10^{20}}$ molecules of $S{O_2}$
b) $3.65 \times {10^{20}}$ molecules of $S{O_3}$
c) 0.0553 g of $S{O_3}$
d) $5.06 \times {10^{ - 3}}$ moles of $S{O_3}$
Answer
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Hint: In this given question, we need to find the number of oxygen atoms. To find the no. of atoms we first have to find the no. of moles of each compound given to us i.e., $S{O_2}$ and $S{O_3}$ . For this question we need to remember the Avogadro’s Number denoted by ${N_A} = 6.022 \times {10^{23}}molecules$
Complete answer:
The Avogadro’s number gives us the no. of molecules present in 1 mole of any compound. It states that 1 mol of any compound will have \[6.022 \times {10^{23}}molecules\] present in it. To find the largest no. of oxygen atoms, we need to find the no. of moles of each compound. More the no. of moles more will be the no. of oxygen atoms.
First, we have been given $3.65 \times {10^{20}}$ molecules of $S{O_2}$ . We know that: $1mol{\text{ }}S{O_2} = 6.022 \times {10^{23}}$
Therefore, $3.65 \times {10^{20}}molecules = \dfrac{{3.65 \times {{10}^{20}}}}{{6.022 \times {{10}^{23}}}} = 6.06 \times {10^{ - 4}}mol$
Since the no. of molecules of $S{O_3}$ and $S{O_2}$, therefore the no. of moles of $S{O_3}$ will be equal to $S{O_2}$ . Hence the no. of moles of $S{O_3}$ will be equal to $6.06 \times {10^{ - 4}}mol$
We are given 0.0553g of $S{O_3}$ . Given that the Molar Mass of $S{O_3}$ is 80.1 g/mol. Therefore the no. of moles of $S{O_3}$ in 0.0553 g $S{O_3}$ $ = \dfrac{{0.0553g}}{{80.1g/mol}} = 6.90 \times {10^{ - 4}}mol$ of $S{O_3}$ .
Out of $S{O_3}$ and $S{O_2}$more no. Oxygen atoms are present in $S{O_3}$ . More the moles of $S{O_3}$ more will be the no. of oxygen atoms. Out of all the four options Option c) has the maximum no. of moles. Hence it will have more no. of oxygen atoms.
Hence the correct answer is Option (C).
Note:
The no. of molecules present in 1 mol of every compound will be the same for all the compounds. This only means that the no. of molecules are the same and nothing else. If the Avogadro’s number is the same it means that it will have the same no. of atoms that comprise the compound and also has the same molecular ratio.
Complete answer:
The Avogadro’s number gives us the no. of molecules present in 1 mole of any compound. It states that 1 mol of any compound will have \[6.022 \times {10^{23}}molecules\] present in it. To find the largest no. of oxygen atoms, we need to find the no. of moles of each compound. More the no. of moles more will be the no. of oxygen atoms.
First, we have been given $3.65 \times {10^{20}}$ molecules of $S{O_2}$ . We know that: $1mol{\text{ }}S{O_2} = 6.022 \times {10^{23}}$
Therefore, $3.65 \times {10^{20}}molecules = \dfrac{{3.65 \times {{10}^{20}}}}{{6.022 \times {{10}^{23}}}} = 6.06 \times {10^{ - 4}}mol$
Since the no. of molecules of $S{O_3}$ and $S{O_2}$, therefore the no. of moles of $S{O_3}$ will be equal to $S{O_2}$ . Hence the no. of moles of $S{O_3}$ will be equal to $6.06 \times {10^{ - 4}}mol$
We are given 0.0553g of $S{O_3}$ . Given that the Molar Mass of $S{O_3}$ is 80.1 g/mol. Therefore the no. of moles of $S{O_3}$ in 0.0553 g $S{O_3}$ $ = \dfrac{{0.0553g}}{{80.1g/mol}} = 6.90 \times {10^{ - 4}}mol$ of $S{O_3}$ .
Out of $S{O_3}$ and $S{O_2}$more no. Oxygen atoms are present in $S{O_3}$ . More the moles of $S{O_3}$ more will be the no. of oxygen atoms. Out of all the four options Option c) has the maximum no. of moles. Hence it will have more no. of oxygen atoms.
Hence the correct answer is Option (C).
Note:
The no. of molecules present in 1 mol of every compound will be the same for all the compounds. This only means that the no. of molecules are the same and nothing else. If the Avogadro’s number is the same it means that it will have the same no. of atoms that comprise the compound and also has the same molecular ratio.
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