When `20ml` of mixture of `O_(2)` and `O_(3)` is heated the volume becomes `29ml` and disappears in alkaline pyragallol solution. What is the volume precent of `O_(2)` in the originl mixture ? .

Options

A. `90%`
B. `10%`
C. `18%`
D. `2%`

Explanation

To solve the problem step by step, we need to analyze the given information and set up the equations based on the reactions involved.

### Step 1: Define Variables
Let:
- \( x \) = volume of \( O_2 \) in the original mixture (in ml)
- \( y \) = volume of \( O_3 \) in the original mixture (in ml)

### Step 2: Set Up the First Equation
From the problem, we know that the total volume of the gas mixture is \( 20 \) ml. Therefore, we can write our first equation as:
\[ x + y = 20 \quad \text{(1)} \]

### Step 3: Understand the Reaction
When \( O_3 \) is heated, it decomposes to form \( O_2 \):
\[ 2 O_3 \rightarrow 3 O_2 \]
From this reaction, we can deduce that for every \( y \) ml of \( O_3 \), it produces \( \frac{3}{2}y \) ml of \( O_2 \).

### Step 4: Set Up the Second Equation
After heating, the total volume of gas becomes \( 29 \) ml. The total volume of gas after the reaction includes the original \( O_2 \) and the \( O_2 \) produced from \( O_3 \):
\[ x + \frac{3}{2}y = 29 \quad \text{(2)} \]

### Step 5: Solve the Equations
Now, we have two equations:
1. \( x + y = 20 \)
2. \( x + \frac{3}{2}y = 29 \)

We can solve these equations simultaneously. From equation (1), we can express \( x \) in terms of \( y \):
\[ x = 20 - y \]

Substituting this expression for \( x \) into equation (2):
\[ (20 - y) + \frac{3}{2}y = 29 \]
Now, simplify this equation:
\[ 20 - y + \frac{3}{2}y = 29 \]
Combine like terms:
\[ 20 + \frac{1}{2}y = 29 \]
Subtract \( 20 \) from both sides:
\[ \frac{1}{2}y = 9 \]
Multiply both sides by \( 2 \):
\[ y = 18 \]

### Step 6: Find \( x \)
Now substitute \( y = 18 \) back into equation (1) to find \( x \):
\[ x + 18 = 20 \]
So,
\[ x = 20 - 18 = 2 \]

### Step 7: Calculate Volume Percent of \( O_2 \)
Now we have:
- Volume of \( O_2 \) in the original mixture, \( x = 2 \) ml
- Total volume of the mixture = 20 ml

The volume percent of \( O_2 \) is calculated as:
\[ \text{Volume percent of } O_2 = \left( \frac{x}{\text{Total Volume}} \right) \times 100 = \left( \frac{2}{20} \right) \times 100 = 10\% \]

### Final Answer
The volume percent of \( O_2 \) in the original mixture is **10%**.

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