The total pressure of a mixture of 6.4 grams of oxygen and 5.6 grams of nitrogen present in a 2 lit vessel is 1200mm. What is the partial pressure of nitrogen in mm?

Options

A. 1200
B. 600
C. 900
D. 200

Answer

To find the partial pressure of nitrogen in a mixture of gases, we can follow these steps:

### Step 1: Calculate the number of moles of each gas.
- **For Oxygen (O₂)**:
- Given mass = 6.4 grams
- Molar mass of O₂ = 32 g/mol (since O has a molar mass of 16 g/mol, and O₂ = 2 × 16)

\[ \text{Number of moles of O₂} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{6.4 \text{ g}}{32 \text{ g/mol}} = 0.2 \text{ moles} \]

- **For Nitrogen (N₂)**:
- Given mass = 5.6 grams
- Molar mass of N₂ = 28 g/mol (since N has a molar mass of 14 g/mol, and N₂ = 2 × 14)

\[ \text{Number of moles of N₂} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{5.6 \text{ g}}{28 \text{ g/mol}} = 0.2 \text{ moles} \]

### Step 2: Calculate the total number of moles in the mixture.
\[ \text{Total moles} = \text{Moles of O₂} + \text{Moles of N₂} = 0.2 + 0.2 = 0.4 \text{ moles} \]

### Step 3: Calculate the mole fraction of nitrogen (N₂).
\[ \text{Mole fraction of N₂} (X_{N₂}) = \frac{\text{Moles of N₂}}{\text{Total moles}} = \frac{0.2}{0.4} = 0.5 \]

### Step 4: Use Dalton's Law of Partial Pressures to find the partial pressure of nitrogen.
- Total pressure (P_total) = 1200 mm
- The formula for partial pressure is:
\[ P_{N₂} = P_{\text{total}} \times X_{N₂} \]
Substituting the values:
\[ P_{N₂} = 1200 \text{ mm} \times 0.5 = 600 \text{ mm} \]

### Final Answer:
The partial pressure of nitrogen (N₂) is **600 mm**.

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