What is the ratio of the number of molecules in 14 g CO to that of 28 g N ?

Options

A. `1.5`
B. 2
C. `3.5`
D. `0.5`

Answer

To solve the question of finding the ratio of the number of molecules in 14 g of carbon monoxide (CO) to that of 28 g of nitrogen (N₂), we will follow these steps:

### Step 1: Determine the molecular mass of carbon monoxide (CO)
- The atomic mass of carbon (C) is 12 g/mol and that of oxygen (O) is 16 g/mol.
- Therefore, the molecular mass of CO = 12 g/mol + 16 g/mol = 28 g/mol.

### Step 2: Determine the molecular mass of nitrogen (N₂)
- The atomic mass of nitrogen (N) is 14 g/mol.
- Therefore, the molecular mass of N₂ = 14 g/mol × 2 = 28 g/mol.

### Step 3: Calculate the number of moles in 14 g of CO
- Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molecular mass (g/mol)}} \]
- For 14 g of CO:
\[ \text{Number of moles of CO} = \frac{14 \text{ g}}{28 \text{ g/mol}} = 0.5 \text{ moles} \]

### Step 4: Calculate the number of molecules in 14 g of CO
- Using Avogadro's number (approximately \(6.022 \times 10^{23}\) molecules/mol):
\[ \text{Number of molecules in 14 g of CO} = 0.5 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mol} = 3.011 \times 10^{23} \text{ molecules} \]

### Step 5: Calculate the number of moles in 28 g of N₂
- For 28 g of N₂:
\[ \text{Number of moles of N₂} = \frac{28 \text{ g}}{28 \text{ g/mol}} = 1 \text{ mole} \]

### Step 6: Calculate the number of molecules in 28 g of N₂
- Using Avogadro's number:
\[ \text{Number of molecules in 28 g of N₂} = 1 \text{ mole} \times 6.022 \times 10^{23} \text{ molecules/mol} = 6.022 \times 10^{23} \text{ molecules} \]

### Step 7: Calculate the ratio of the number of molecules
- The ratio of the number of molecules in 14 g of CO to that in 28 g of N₂ is:
\[ \text{Ratio} = \frac{3.011 \times 10^{23} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules}} = \frac{3.011}{6.022} = \frac{1}{2} = 0.5 \]

### Final Answer
The ratio of the number of molecules in 14 g of CO to that of 28 g of N₂ is **0.5**.

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