A gasesous mixture contains 56 g`N_(2)`,44 g `CO_(2)` and 16 g `CH_(4)`. The total pressure of the mixture is 720 mm Hg. What is the partial pressure of `CH_(4)` ?

Explanation

To find the partial pressure of \( CH_4 \) in the given gaseous mixture, we can follow these steps:

### Step 1: Calculate the number of moles of each gas
We will use the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

1. **For \( N_2 \)**:
- Mass = 56 g
- Molar mass of \( N_2 \) = 28 g/mol
\[ n_{N_2} = \frac{56 \, \text{g}}{28 \, \text{g/mol}} = 2 \, \text{moles} \]

2. **For \( CO_2 \)**:
- Mass = 44 g
- Molar mass of \( CO_2 \) = 44 g/mol
\[ n_{CO_2} = \frac{44 \, \text{g}}{44 \, \text{g/mol}} = 1 \, \text{mole} \]

3. **For \( CH_4 \)**:
- Mass = 16 g
- Molar mass of \( CH_4 \) = 16 g/mol
\[ n_{CH_4} = \frac{16 \, \text{g}}{16 \, \text{g/mol}} = 1 \, \text{mole} \]

### Step 2: Calculate the total number of moles in the mixture
\[ n_{total} = n_{N_2} + n_{CO_2} + n_{CH_4} = 2 + 1 + 1 = 4 \, \text{moles} \]

### Step 3: Calculate the mole fraction of \( CH_4 \)
The mole fraction \( X_{CH_4} \) is given by:
\[ X_{CH_4} = \frac{n_{CH_4}}{n_{total}} = \frac{1}{4} = 0.25 \]

### Step 4: Use Dalton's Law of Partial Pressures to find the partial pressure of \( CH_4 \)
According to Dalton's Law:
\[ P_{CH_4} = X_{CH_4} \times P_{total} \]
Where \( P_{total} = 720 \, \text{mm Hg} \):
\[ P_{CH_4} = 0.25 \times 720 \, \text{mm Hg} = 180 \, \text{mm Hg} \]

### Final Answer
The partial pressure of \( CH_4 \) is \( 180 \, \text{mm Hg} \).

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