Question

The largest number of molecules is in:

• A. 36 g of water
• B. 28 g of \( \text{CO}_2 \)
• C. 46 g of \( \text{CH}_3\text{OH} \)
• D. 58 g of \( \text{N}_2\text{O}_5 \)

Hint:

To find the number of molecules, divide the mass by the molar mass to get the number of moles, then multiply by Avogadro’s number.

Answer 1

The Correct Option is A

Step 1: Calculate the number of moles in each substance.
To find the largest number of molecules, we need to calculate the number of moles in each substance. The number of molecules in one mole of a substance is Avogadro’s number \( 6.022 \times 10^{23} \). - For water (H₂O), molar mass = 18 g/mol. So, in 36 g of water: \[ \text{Number of moles} = \frac{36}{18} = 2 \, \text{moles} \] Thus, the number of molecules in 36 g of water is: \[ 2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24} \, \text{molecules} \] - For \( \text{CO}_2 \), molar mass = 44 g/mol. So, in 28 g of \( \text{CO}_2 \): \[ \text{Number of moles} = \frac{28}{44} = 0.636 \, \text{moles} \] Thus, the number of molecules in 28 g of \( \text{CO}_2 \) is: \[ 0.636 \times 6.022 \times 10^{23} = 3.83 \times 10^{23} \, \text{molecules} \] - For \( \text{CH}_3\text{OH} \), molar mass = 46 g/mol. So, in 46 g of \( \text{CH}_3\text{OH} \): \[ \text{Number of moles} = \frac{46}{46} = 1 \, \text{mole} \] Thus, the number of molecules in 46 g of \( \text{CH}_3\text{OH} \) is: \[ 1 \times 6.022 \times 10^{23} = 6.022 \times 10^{23} \, \text{molecules} \] - For \( \text{N}_2\text{O}_5 \), molar mass = 108 g/mol. So, in 58 g of \( \text{N}_2\text{O}_5 \): \[ \text{Number of moles} = \frac{58}{108} = 0.537 \, \text{moles} \] Thus, the number of molecules in 58 g of \( \text{N}_2\text{O}_5 \) is: \[ 0.537 \times 6.022 \times 10^{23} = 3.23 \times 10^{23} \, \text{molecules} \]

Step 2: Conclusion.
The largest number of molecules is in 36 g of water, which corresponds to option (1).