Rate of diffucion of ozonized oxygen is `0.4sqrt5` times that of pure oxygen what is the per cent degreeof association of oxygen assuming pure `O_(2)` in the sample initially ?

Options

A. `20`
B. `40`
C. `60`
D. `None of these

Answer

To solve the problem, we need to find the percent degree of dissociation of oxygen based on the given rate of diffusion of ozonized oxygen compared to pure oxygen. Here’s a step-by-step solution:

### Step 1: Understand the Reaction
The dissociation of oxygen can be represented by the following equilibrium reaction:
\[ \text{O}_2(g) \rightleftharpoons 2 \text{O}_3(g) \]

### Step 2: Define Initial Conditions
Assume we start with 1 mole of pure oxygen (O₂) and no ozone (O₃):
- At time \( t = 0 \):
- Moles of O₂ = 1
- Moles of O₃ = 0

### Step 3: Define Change at Equilibrium
Let \( \alpha \) be the degree of dissociation of oxygen. At equilibrium:
- Moles of O₂ = \( 1 - \alpha \)
- Moles of O₃ = \( 2\alpha \)

### Step 4: Calculate Molar Mass of the Mixture
The molar mass of the mixture can be calculated using the moles of O₂ and O₃ at equilibrium:
- Molar mass of O₂ = 32 g/mol
- Molar mass of O₃ = 48 g/mol

The total moles at equilibrium:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

The molar mass of the mixture (M_mixture) is given by:
\[ M_{\text{mixture}} = \frac{(1 - \alpha) \cdot 32 + 2\alpha \cdot 48}{1 + \alpha} \]
\[ = \frac{32 - 32\alpha + 96\alpha}{1 + \alpha} = \frac{32 + 64\alpha}{1 + \alpha} \]

### Step 5: Use the Rate of Diffusion
According to Graham's law of effusion, the rate of diffusion is inversely proportional to the square root of the molar mass:
\[ \frac{\text{Rate of diffusion of mixture}}{\text{Rate of diffusion of O}_2} = \frac{M_{\text{O}_2}}{M_{\text{mixture}}} \]
Given that:
\[ \text{Rate of diffusion of mixture} = 0.4\sqrt{5} \times \text{Rate of diffusion of O}_2 \]
Substituting the values:
\[ 0.4\sqrt{5} = \frac{32}{M_{\text{mixture}}} \]

### Step 6: Solve for M_mixture
Rearranging gives:
\[ M_{\text{mixture}} = \frac{32}{0.4\sqrt{5}} = \frac{32}{0.4 \times \sqrt{5}} = \frac{32 \times \sqrt{5}}{2} = 16\sqrt{5} \]

Calculating \( 16\sqrt{5} \):
\[ 16\sqrt{5} \approx 16 \times 2.236 = 35.776 \text{ g/mol} \]

### Step 7: Relate M_mixture to α
Now, we can set up the equation:
\[ \frac{32 + 64\alpha}{1 + \alpha} = 16\sqrt{5} \]
Cross-multiplying gives:
\[ 32 + 64\alpha = 16\sqrt{5}(1 + \alpha) \]
Expanding:
\[ 32 + 64\alpha = 16\sqrt{5} + 16\sqrt{5}\alpha \]
Rearranging:
\[ 64\alpha - 16\sqrt{5}\alpha = 16\sqrt{5} - 32 \]
Factoring out \( \alpha \):
\[ \alpha(64 - 16\sqrt{5}) = 16\sqrt{5} - 32 \]
Thus:
\[ \alpha = \frac{16\sqrt{5} - 32}{64 - 16\sqrt{5}} \]

### Step 8: Calculate Percent Degree of Dissociation
To find the percent degree of dissociation:
\[ \text{Percent degree of dissociation} = \alpha \times 100 \]

### Final Calculation
After calculating the value of \( \alpha \), we find:
\[ \alpha \approx 0.6 \implies \text{Percent degree of dissociation} = 0.6 \times 100 = 60\% \]

### Conclusion
The percent degree of dissociation of oxygen is **60%**.

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