A mixture of `CH_(4), N_(2)` and `O_(2)` is enclosed in a container of 1 litre capacity at `0^(@)C`. Total pressure of gaseous mixture is 2660 mm Hg. If the ratio of partial pressures of the gases is 1 : 4 : 2 respectively, the number of moles of oxygen present in the vessel is:

Options

A. `(1)/(22.4)`
B. 1
C. 0.1
D. none of these

Explanation

To find the number of moles of oxygen (O₂) present in a mixture of gases (CH₄, N₂, and O₂) under the given conditions, we can follow these steps:

### Step 1: Understand the Given Data
- Total pressure (P_total) = 2660 mm Hg
- Volume of the container (V) = 1 L
- Temperature (T) = 0°C = 273 K
- Ratio of partial pressures of CH₄ : N₂ : O₂ = 1 : 4 : 2

### Step 2: Express Partial Pressures in Terms of a Variable
Let the partial pressures of the gases be:
- P(CH₄) = 1x
- P(N₂) = 4x
- P(O₂) = 2x

### Step 3: Write the Equation for Total Pressure
The total pressure is the sum of the partial pressures:
\[ P_{total} = P(CH₄) + P(N₂) + P(O₂) \]
Substituting the expressions for partial pressures:
\[ 2660 \, \text{mm Hg} = 1x + 4x + 2x \]
\[ 2660 = 7x \]

### Step 4: Solve for x
To find the value of x:
\[ x = \frac{2660}{7} \]
Calculating this gives:
\[ x = 380 \, \text{mm Hg} \]

### Step 5: Find the Partial Pressure of O₂
Now, we can find the partial pressure of O₂:
\[ P(O₂) = 2x = 2 \times 380 = 760 \, \text{mm Hg} \]

### Step 6: Use Ideal Gas Law to Find Moles of O₂
Using the ideal gas law, \( PV = nRT \):
- P = 760 mm Hg (which is equivalent to 1 atm)
- V = 1 L
- R = 0.0821 L·atm/(K·mol)
- T = 273 K

Rearranging the ideal gas law to find n (number of moles):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(1 \, \text{atm})(1 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)})(273 \, \text{K})} \]

### Step 7: Calculate n
Calculating the denominator:
\[ n = \frac{1}{0.0821 \times 273} \]
\[ n = \frac{1}{22.4143} \approx 0.0446 \, \text{mol} \]

### Conclusion
Thus, the number of moles of oxygen (O₂) present in the vessel is approximately **0.0446 moles**.

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