A given volume of ozonised oxygen (containing 60% oxygen by volume ) required 220 sec to effuse while an equal volume of oxygen took 200 sec only under identical conditions. If density of `O_(2)` is 1.6 g/L then find density of `O_(3)`.

Options

A. 1.963 g/L
B. 2.16 g/L
C. 3.28 g/L
D. 2.24 g/L

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Understand the problem
We have a mixture of ozonised oxygen (60% O2 and 40% O3) and we need to find the density of ozone (O3). We know the effusion times for both the ozonised oxygen and pure oxygen, as well as the density of oxygen.

### Step 2: Use Graham's law of effusion
According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The formula can be expressed as:

\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]

Where:
- \( R_1 \) and \( R_2 \) are the rates of effusion of the two gases.
- \( M_1 \) and \( M_2 \) are the molar masses of the two gases.

### Step 3: Calculate the rates of effusion
Given:
- Time for ozonised oxygen (T1) = 220 seconds
- Time for oxygen (T2) = 200 seconds

The rates of effusion can be expressed as:
\[ R_1 = \frac{V}{T_1} \quad \text{and} \quad R_2 = \frac{V}{T_2} \]

Thus, we can write:
\[ \frac{R_1}{R_2} = \frac{T_2}{T_1} = \frac{200}{220} = \frac{10}{11} \]

### Step 4: Relate the rates to molar masses
Let \( M_{O2} \) be the molar mass of oxygen and \( M_{O3} \) be the molar mass of ozone. We know that:

\[ \frac{10}{11} = \sqrt{\frac{M_{O3}}{M_{O2}}} \]

Squaring both sides gives:
\[ \left(\frac{10}{11}\right)^2 = \frac{M_{O3}}{M_{O2}} \]

### Step 5: Calculate the molar mass of O3
Using the molar mass of O2 (approximately 32 g/mol):
\[ \frac{100}{121} = \frac{M_{O3}}{32} \]

Cross-multiplying gives:
\[ M_{O3} = 32 \times \frac{100}{121} \approx 26.45 \, \text{g/mol} \]

### Step 6: Calculate the density of the ozonised oxygen mixture
We know the density of the mixture (D_mixture) is given as:
\[ D_{mixture} = \frac{mass_{O2} + mass_{O3}}{volume} \]

Given that the mixture is 60% O2 and 40% O3 by volume, we can express the mass of each component:
- Volume of O2 = 60 L (assuming total volume = 100 L)
- Volume of O3 = 40 L

Using the density of O2 (1.6 g/L):
\[ mass_{O2} = 1.6 \, \text{g/L} \times 60 \, \text{L} = 96 \, \text{g} \]

Let \( D_{O3} \) be the density of O3. Then:
\[ mass_{O3} = D_{O3} \times 40 \]

The total mass of the mixture is:
\[ mass_{mixture} = mass_{O2} + mass_{O3} = 96 + D_{O3} \times 40 \]

### Step 7: Use the density of the mixture to find D3
Using the previously calculated density of the mixture (1.936 g/L):
\[ D_{mixture} = \frac{96 + D_{O3} \times 40}{100} \]

Setting this equal to the known density of the mixture:
\[ 1.936 = \frac{96 + D_{O3} \times 40}{100} \]

Cross-multiplying gives:
\[ 193.6 = 96 + D_{O3} \times 40 \]

Solving for \( D_{O3} \):
\[ D_{O3} \times 40 = 193.6 - 96 \]
\[ D_{O3} \times 40 = 97.6 \]
\[ D_{O3} = \frac{97.6}{40} \approx 2.44 \, \text{g/L} \]

### Final Answer
The density of ozone (O3) is approximately **2.44 g/L**.

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