Question

Which of the following compounds have the largest number of molecules?

• A. $36\mathrm{g}$ of water
• B. $28\mathrm{g}$ of $\mathrm{CO}$
• C. $46\mathrm{g}$ of ethyl alcohol
• D. $54\mathrm{g}$ of nitrogen pentaoxide $\left({\mathrm{N}}_2{\mathrm{O}}_5\right)$

Answer 1

The correct option is A

$36\mathrm{g}$ of water

Explanation for correct option:

A. $36\mathrm{g}$ of water

Step 1: One mole:

• “The mole, symbol mol, is the SI unit of amount of substance. One mole contains exactly $6.02214076\times {10}^{23}$ elementary entities.”
• “This number is the fixed numerical value of the Avogadro constant, NA, when expressed in the unit ${\mathrm{mol}}^{-1}$ and is called the Avogadro number.”

Step 2: Number of moles:

• The number of moles is defined as the given mass over the molar mass of the substance.
• It is expressed as:$\mathrm{N}=\frac{\mathrm{m}}{\mathrm{M}}$, where m is the given mass while M is the molar mass.

Step 3: Calculation of the number of molecules in $36\mathrm{g}$ of water:

• Given mass, m= $36\mathrm{g}$
• Molar mass, M = $18\mathrm{g}{\mathrm{mol}}^{-1}$
• The number of moles, N will be:
• $\mathrm{N}=\frac{36\mathrm{g}}{18\mathrm{g}{\mathrm{mol}}^{-1}}=2\mathrm{mol}$
• $1\mathrm{mol}=6.02214076\times {10}^{23}\mathrm{molecules}$
• Thus, For 2 moles, $2\mathrm{mol}=2\times 6.02214076\times {10}^{23}\mathrm{molecules}$

Explanation for incorrect options:

B. $28\mathrm{g}$ of $\mathrm{CO}$

Calculation of the number of molecules in $28\mathrm{g}$ of $\mathrm{CO}$:

• Given mass, m= $28\mathrm{g}$
• Molar mass, M = $28.01\mathrm{g}{\mathrm{mol}}^{-1}$
• The number of moles, N will be:
• $\mathrm{N}=\frac{28\mathrm{g}}{28.01\mathrm{g}{\mathrm{mol}}^{-1}}=0.01\mathrm{mol}$
• $1\mathrm{mol}=6.02214076\times {10}^{23}\mathrm{molecules}$
• Thus, For 0.01 moles, $0.01\mathrm{mol}=0.01\times 6.02214076\times {10}^{23}\mathrm{molecules}$.

C. $46\mathrm{g}$ of ethyl alcohol$\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right)$:

Calculation of the number of molecules in $46\mathrm{g}$ of ethyl alcohol$\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right)$:

• Given mass, m= $46\mathrm{g}$
• Molar mass, M = $46.07\mathrm{g}{\mathrm{mol}}^{-1}$
• The number of moles, N will be:
• $\mathrm{N}=\frac{46\mathrm{g}}{46.07\mathrm{g}{\mathrm{mol}}^{-1}}=0.07\mathrm{mol}$
• $1\mathrm{mol}=6.02214076\times {10}^{23}\mathrm{molecules}$
• Thus, For 0.07 moles, $0.07\mathrm{mol}=0.07\times 6.02214076\times {10}^{23}\mathrm{molecules}$.

D. $54\mathrm{g}$ of nitrogen pentaoxide $\left({\mathrm{N}}_2{\mathrm{O}}_5\right)$:

Calculation of the number of molecules in $54\mathrm{g}$ of nitrogen pentaoxide $\left({\mathrm{N}}_2{\mathrm{O}}_5\right)$:

• Given mass, m= $54\mathrm{g}$
• Molar mass, M = $108.01\mathrm{g}{\mathrm{mol}}^{-1}$
• The number of moles, N will be:
• $\mathrm{N}=\frac{54\mathrm{g}}{108.01\mathrm{g}{\mathrm{mol}}^{-1}}=0.5\mathrm{mol}$
• $1\mathrm{mol}=6.02214076\times {10}^{23}\mathrm{molecules}$
• Thus, For 0.5 moles, $0.5\mathrm{mol}=0.5\times 6.02214076\times {10}^{23}\mathrm{molecules}$.

Hence, the largest number of molecules are present in $36\mathrm{g}$ of water i.e.$2\times 6.02214076\times {10}^{23}\mathrm{molecules}$. Thus option A is correct.