`600 mL` of ozonised oxygen at `STP` were found to weigh one gram. What is the volume of ozone in the ozonised oxygen?

Options

A. `200 mL`
B. `150 mL`
C. `100 mL`
D. `50 mL`

Explanation

To solve the problem of finding the volume of ozone in ozonised oxygen, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = volume of ozone (O₃) in mL.
- Volume of oxygen (O₂) = \( 600 - x \) mL.

### Step 2: Understand the Molar Volume at STP
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters (or 22,400 mL).

### Step 3: Calculate the Mass of Ozone
The molar mass of ozone (O₃) is 48 g/mol. The mass of \( x \) mL of ozone can be calculated using the formula:
\[ \text{Mass of O₃} = \left(\frac{x}{22400}\right) \times 48 \]

### Step 4: Calculate the Mass of Oxygen
The molar mass of oxygen (O₂) is 32 g/mol. The mass of \( 600 - x \) mL of oxygen can be calculated as:
\[ \text{Mass of O₂} = \left(\frac{600 - x}{22400}\right) \times 32 \]

### Step 5: Set Up the Equation
According to the problem, the total mass of the ozonised oxygen is 1 gram. Therefore, we can write the equation:
\[ \left(\frac{x}{22400}\right) \times 48 + \left(\frac{600 - x}{22400}\right) \times 32 = 1 \]

### Step 6: Simplify the Equation
Multiply through by 22400 to eliminate the denominator:
\[ 48x + 32(600 - x) = 22400 \]

### Step 7: Expand and Rearrange
Expanding the equation gives:
\[ 48x + 19200 - 32x = 22400 \]
Combine like terms:
\[ 16x + 19200 = 22400 \]

### Step 8: Solve for \( x \)
Subtract 19200 from both sides:
\[ 16x = 3200 \]
Now, divide by 16:
\[ x = 200 \text{ mL} \]

### Conclusion
The volume of ozone in the ozonised oxygen is **200 mL**.

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