
From the following the largest number of atoms are in
A. 36 g ${{H}_{2}}O$
B. 28 g of CO
C. 46 g of ${{C}_{2}}{{H}_{5}}OH$
D. 54 g of ${{N}_{2}}{{O}_{5}}$
Answer
570.3k+ views
Hint: One mole of a compound contains Avogadro number of atoms in it. There is a formula to calculate the number of moles and it is as follows.
Number of moles of a compound = $\dfrac{W}{MW}$
W = Weight of the given compound
MW = molecular weight of the given compound
Complete step by step solution:
- In the question it is given that which compound contains a large number of atoms among the given options.
- Coming to the given options, Option A, 36 g of water.
- Number of moles of Water in 36 g of water is
\[\begin{align}
&\implies \dfrac{W}{MW} \\
&\implies \dfrac{36}{18} \\
&\therefore 2mol \\
\end{align}\]
- 2 moles of water contains $2\times 6.023\times {{10}^{23}}$ number of atoms.
- Coming to option B, 28 g of carbon monoxide.
- Number of moles of CO in 28 g of CO is
\[\begin{align}
&\implies \dfrac{W}{MW} \\
&\implies \dfrac{28}{28} \\
&\therefore 1mol \\
\end{align}\]
- 1 mole of CO contains $1\times 6.023\times {{10}^{23}}$ number of atoms.
- Coming to option C, 46 g of ethyl alcohol.
- Number of moles of ethyl alcohol in 46 g of ethyl alcohol is
\[\begin{align}
&\implies \dfrac{W}{MW} \\
&\implies \dfrac{46}{46} \\
&\therefore 1mol \\
\end{align}\]
- 1 mole of ethyl alcohol contains $1\times 6.023\times {{10}^{23}}$ number of atoms.
- Coming to option D, 56 g of ${{N}_{2}}{{O}_{5}}$ .
- Number of moles of ${{N}_{2}}{{O}_{5}}$ in 56 g of ${{N}_{2}}{{O}_{5}}$ is
\[\begin{align}
&\implies \dfrac{W}{MW} \\
&\implies \dfrac{54}{108} \\
&\therefore 0.5mol \\
\end{align}\]
- 1 mole of ${{N}_{2}}{{O}_{5}}$ contains $0.5\times 6.023\times {{10}^{23}}$ number of atoms.
- Therefore 36 g of water contains the highest number of atoms among the following options.
So, the correct option is A.
Note: To know about the number of atoms present in the compound first we should calculate the number of moles of the given compound. Later we should convert the number of moles of the given compound into the number of atoms by using Avogadro number.
Number of moles of a compound = $\dfrac{W}{MW}$
W = Weight of the given compound
MW = molecular weight of the given compound
Complete step by step solution:
- In the question it is given that which compound contains a large number of atoms among the given options.
- Coming to the given options, Option A, 36 g of water.
- Number of moles of Water in 36 g of water is
\[\begin{align}
&\implies \dfrac{W}{MW} \\
&\implies \dfrac{36}{18} \\
&\therefore 2mol \\
\end{align}\]
- 2 moles of water contains $2\times 6.023\times {{10}^{23}}$ number of atoms.
- Coming to option B, 28 g of carbon monoxide.
- Number of moles of CO in 28 g of CO is
\[\begin{align}
&\implies \dfrac{W}{MW} \\
&\implies \dfrac{28}{28} \\
&\therefore 1mol \\
\end{align}\]
- 1 mole of CO contains $1\times 6.023\times {{10}^{23}}$ number of atoms.
- Coming to option C, 46 g of ethyl alcohol.
- Number of moles of ethyl alcohol in 46 g of ethyl alcohol is
\[\begin{align}
&\implies \dfrac{W}{MW} \\
&\implies \dfrac{46}{46} \\
&\therefore 1mol \\
\end{align}\]
- 1 mole of ethyl alcohol contains $1\times 6.023\times {{10}^{23}}$ number of atoms.
- Coming to option D, 56 g of ${{N}_{2}}{{O}_{5}}$ .
- Number of moles of ${{N}_{2}}{{O}_{5}}$ in 56 g of ${{N}_{2}}{{O}_{5}}$ is
\[\begin{align}
&\implies \dfrac{W}{MW} \\
&\implies \dfrac{54}{108} \\
&\therefore 0.5mol \\
\end{align}\]
- 1 mole of ${{N}_{2}}{{O}_{5}}$ contains $0.5\times 6.023\times {{10}^{23}}$ number of atoms.
- Therefore 36 g of water contains the highest number of atoms among the following options.
So, the correct option is A.
Note: To know about the number of atoms present in the compound first we should calculate the number of moles of the given compound. Later we should convert the number of moles of the given compound into the number of atoms by using Avogadro number.
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