On subjecting 10ml mixture of `N_(2)` and CO to repeated electric spark to form `CO_(2)` and NO, 7 ml of `O_(2)` was required for combustion. What was the mole precent of CO in the mixture ? (All volumes were measured under identical conditions)

Options

A. 4
B. 6
C. 40
D. 60

Answer

To solve the problem step-by-step, we can follow these steps:

### Step 1: Define the variables
Let the moles of carbon monoxide (CO) in the mixture be \( X \). Since the total volume of the gas mixture is 10 ml, the moles of nitrogen (N₂) can be expressed as \( 10 - X \).

### Step 2: Write the reactions
The reactions that take place during the combustion are:
1. For carbon monoxide:
\[ \text{CO} + \frac{1}{2} \text{O}_2 \rightarrow \text{CO}_2 \]
From this reaction, we can see that 1 mole of CO reacts with 0.5 moles of O₂.

2. For nitrogen:
\[ \text{N}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{2NO} \]
Here, 1 mole of N₂ reacts with 1 mole of O₂ to produce 2 moles of NO.

### Step 3: Calculate the oxygen required
From the above reactions:
- The amount of oxygen required for \( X \) moles of CO is \( \frac{X}{2} \) ml.
- The amount of oxygen required for \( 10 - X \) moles of N₂ is \( 10 - X \) ml.

### Step 4: Set up the equation based on the total oxygen used
According to the problem, the total volume of oxygen used is 7 ml. Therefore, we can set up the equation:
\[ \frac{X}{2} + (10 - X) = 7 \]

### Step 5: Solve the equation
Now, simplify the equation:
\[ \frac{X}{2} + 10 - X = 7 \]
Combine like terms:
\[ 10 - \frac{X}{2} = 7 \]
Subtract 10 from both sides:
\[ -\frac{X}{2} = 7 - 10 \]
\[ -\frac{X}{2} = -3 \]
Multiply both sides by -2:
\[ X = 6 \]

### Step 6: Calculate the mole percent of CO
Now that we have \( X = 6 \) (the moles of CO), we can calculate the mole percent of CO in the mixture:
\[ \text{Mole percent of CO} = \left( \frac{X}{\text{Total moles}} \right) \times 100 = \left( \frac{6}{10} \right) \times 100 = 60\% \]

### Final Answer
The mole percent of CO in the mixture is **60%**.

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