In which of the following pairs do 1 g of each have an equal number of molecules .

Options

A. `N_(2)O "and" CO`
B. `N_(2) "and" C_(3)O_(2)`
C. `N_(2) "and" CO`
D. `N_(2)O "and" C_(2)O`

Explanation

To solve the problem of determining which pair of substances contains an equal number of molecules when 1 g of each is taken, we need to follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Moles**:
The number of molecules in a substance can be calculated using the formula:
\[ \text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's number} \]
where the number of moles is given by:
\[ n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

2. **Calculate Molar Masses**:
- For **N2O**:
- Nitrogen (N) = 14 g/mol, Oxygen (O) = 16 g/mol
- Molar mass of N2O = \(2 \times 14 + 16 = 44 \, \text{g/mol}\)

- For **CO**:
- Carbon (C) = 12 g/mol, Oxygen (O) = 16 g/mol
- Molar mass of CO = \(12 + 16 = 28 \, \text{g/mol}\)

3. **Calculate the Number of Moles for Each Substance**:
- For **1 g of N2O**:
\[ n_{N2O} = \frac{1 \, \text{g}}{44 \, \text{g/mol}} \approx 0.02273 \, \text{mol} \]
- For **1 g of CO**:
\[ n_{CO} = \frac{1 \, \text{g}}{28 \, \text{g/mol}} \approx 0.03571 \, \text{mol} \]

4. **Compare the Number of Moles**:
Since \( n_{N2O} \neq n_{CO} \), 1 g of N2O and 1 g of CO do not have the same number of molecules.

5. **Repeat for Other Pairs**:
- For **N2** and **C3O2**:
- Molar mass of N2 = \(2 \times 14 = 28 \, \text{g/mol}\)
- Molar mass of C3O2 = \(3 \times 12 + 2 \times 16 = 36 + 32 = 68 \, \text{g/mol}\)
- Calculate moles:
\[ n_{N2} = \frac{1 \, \text{g}}{28 \, \text{g/mol}} \approx 0.03571 \, \text{mol} \]
\[ n_{C3O2} = \frac{1 \, \text{g}}{68 \, \text{g/mol}} \approx 0.01471 \, \text{mol} \]
- Since \( n_{N2} \neq n_{C3O2} \), they do not have the same number of molecules.

6. **Final Pair Comparison (N2 and CO)**:
- Molar mass of N2 = 28 g/mol (as calculated before).
- Molar mass of CO = 28 g/mol (as calculated before).
- Calculate moles:
\[ n_{N2} = \frac{1 \, \text{g}}{28 \, \text{g/mol}} \approx 0.03571 \, \text{mol} \]
\[ n_{CO} = \frac{1 \, \text{g}}{28 \, \text{g/mol}} \approx 0.03571 \, \text{mol} \]
- Since \( n_{N2} = n_{CO} \), 1 g of N2 and 1 g of CO have the same number of molecules.

### Conclusion:
The pair that has an equal number of molecules when 1 g of each is taken is **N2 and CO**.