The weight of `1 L` of ozonised oxygen at `STP` was found to be `1.5 g`. When `100 mL` of this mixture at `STP` was treated with turpentine oil, the volume was reduced to `90 mL`. The molecular weight of ozone is

Options

A. 49
B. 47
C. 46
D. 47.9

Explanation

To find the molecular weight of ozone from the given data, we will follow these steps:

### Step 1: Calculate the number of moles of the gas mixture
Given that the weight of `1 L` of ozonised oxygen at `STP` is `1.5 g`, we can find the number of moles in `1 L` of the gas mixture.

At STP (Standard Temperature and Pressure), `1 L` of gas corresponds to `22.4 L` of gas, which contains `1 mole` of gas.

Using the formula:
\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]
We can find the number of moles in `1 L` of the mixture:
\[ \text{Number of moles} = \frac{1.5 \, \text{g}}{M} \]
Where \( M \) is the molar mass of the mixture.

### Step 2: Calculate the molar mass of the gas mixture
We know that `1 L` of the gas mixture corresponds to `0.0446 moles` at STP. Therefore, we can equate:
\[ \frac{1.5 \, \text{g}}{M} = 0.0446 \]
Rearranging gives:
\[ M = \frac{1.5 \, \text{g}}{0.0446} \approx 33.63 \, \text{g/mol} \]

### Step 3: Analyze the volume change when treated with turpentine oil
When `100 mL` of the mixture is treated with turpentine oil, the volume reduces to `90 mL`. This means that `10 mL` of the mixture corresponds to ozone (O₃), while the remaining `90 mL` corresponds to oxygen (O₂).

### Step 4: Calculate the moles of ozone and oxygen
Since the volume is directly proportional to the number of moles, we can calculate the moles of ozone and oxygen:
- Moles of ozone (from `10 mL`):
\[ \text{Moles of O}_3 = \frac{10 \, \text{mL}}{100 \, \text{mL}} \times 0.0446 \, \text{moles} = 0.00446 \, \text{moles} \]
- Moles of oxygen (from `90 mL`):
\[ \text{Moles of O}_2 = \frac{90 \, \text{mL}}{100 \, \text{mL}} \times 0.0446 \, \text{moles} = 0.04014 \, \text{moles} \]

### Step 5: Set up the equation for the molar mass of the mixture
Using the moles calculated, we can set up the equation for the average molar mass of the mixture:
\[ \frac{(M_{O_3} \times \text{moles of O}_3) + (M_{O_2} \times \text{moles of O}_2)}{\text{Total moles}} = M_{\text{mixture}} \]
Where:
- \( M_{O_3} \) is the molar mass of ozone (unknown)
- \( M_{O_2} = 32 \, \text{g/mol} \)
- Total moles = \( 0.00446 + 0.04014 = 0.0446 \, \text{moles} \)

Substituting the values:
\[ \frac{(M_{O_3} \times 0.00446) + (32 \times 0.04014)}{0.0446} = 33.63 \]

### Step 6: Solve for the molar mass of ozone
Rearranging gives:
\[ M_{O_3} \times 0.00446 + 1.28448 = 33.63 \times 0.0446 \]
Calculating \( 33.63 \times 0.0446 \):
\[ 1.498058 \]
Thus:
\[ M_{O_3} \times 0.00446 = 1.498058 - 1.28448 \]
\[ M_{O_3} \times 0.00446 = 0.213578 \]
\[ M_{O_3} = \frac{0.213578}{0.00446} \approx 47.9 \, \text{g/mol} \]

### Conclusion
The molecular weight of ozone is approximately \( 48 \, \text{g/mol} \).

---