Rate of diffusion of ozonized oxygen is `0.4 sqrt(5)` times that of pure oxygen. What is the percent degree of association of oxygen assuming pure `O_(2)` in the sample initially ?

Options

A. a. 20
B. b. 40
C. c. 60
D. d. 30

Answer

To solve the problem step by step, we will follow the information given in the question and apply the principles of diffusion and chemical equilibrium.

### Step 1: Understanding the Problem
We are given that the rate of diffusion of ozonized oxygen (O3) is \(0.4 \sqrt{5}\) times that of pure oxygen (O2). We need to find the percent degree of association of oxygen, assuming that we start with pure O2.

### Step 2: Setting Up the Relationship
According to Graham's law of diffusion, the rate of diffusion is inversely proportional to the square root of the molar mass of the gas. Therefore, we can express the relationship as follows:

\[ \frac{\text{Rate of diffusion of O3}}{\text{Rate of diffusion of O2}} = \frac{\sqrt{M_{O2}}}{\sqrt{M_{O3}}} \]

Where \(M_{O2}\) is the molar mass of oxygen (32 g/mol) and \(M_{O3}\) is the molar mass of ozone (48 g/mol).

### Step 3: Expressing the Rates
Let the rate of diffusion of pure O2 be \(R_{O2}\). Then the rate of diffusion of ozonized oxygen is given by:

\[ R_{O3} = 0.4 \sqrt{5} \cdot R_{O2} \]

### Step 4: Applying Graham's Law
Using Graham's law, we can set up the equation:

\[ \frac{0.4 \sqrt{5} \cdot R_{O2}}{R_{O2}} = \frac{\sqrt{32}}{\sqrt{48}} \]

This simplifies to:

\[ 0.4 \sqrt{5} = \frac{\sqrt{32}}{\sqrt{48}} \]

### Step 5: Simplifying the Right Side
Calculating the right side:

\[ \frac{\sqrt{32}}{\sqrt{48}} = \frac{4\sqrt{2}}{4\sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}} \]

### Step 6: Squaring Both Sides
Now we square both sides to eliminate the square roots:

\[ (0.4 \sqrt{5})^2 = \frac{2}{3} \]

Calculating the left side:

\[ 0.16 \cdot 5 = 0.8 \]

So we have:

\[ 0.8 = \frac{2}{3} \]

### Step 7: Finding the Mass of the Mixture
Now we need to find the mass of the mixture. The mass of the mixture can be expressed as:

\[ \text{Mass of mixture} = \text{mass of pure O2} + \text{mass of O3} \]

Let \(n\) be the initial moles of O2. At equilibrium, if \( \alpha \) is the degree of association, the moles of O2 remaining will be \(1 - \alpha\) and the moles of O3 formed will be \(\frac{3}{2} \alpha\).

### Step 8: Setting Up the Mass Equation
The mass of pure O2 is:

\[ \text{Mass of O2} = 32(1 - \alpha) \]

The mass of O3 is:

\[ \text{Mass of O3} = 48 \cdot \frac{3}{2} \alpha = 72 \alpha \]

Thus, the total mass of the mixture is:

\[ \text{Mass of mixture} = 32(1 - \alpha) + 72 \alpha = 32 - 32\alpha + 72\alpha = 32 + 40\alpha \]

### Step 9: Setting Up the Final Equation
From Graham's law, we have:

\[ \frac{32}{32 + 40\alpha} = \frac{1 - \alpha + \frac{3}{2}\alpha}{1} \]

This simplifies to:

\[ \frac{32}{32 + 40\alpha} = 1 - \frac{1}{2}\alpha \]

### Step 10: Cross-Multiplying and Solving for Alpha
Cross-multiplying gives:

\[ 32(1 - \frac{1}{2}\alpha) = 32 + 40\alpha \]

Expanding and rearranging:

\[ 32 - 16\alpha = 32 + 40\alpha \]

Combining like terms:

\[ -16\alpha - 40\alpha = 0 \]

This gives:

\[ -56\alpha = 0 \implies \alpha = \frac{32}{56} = \frac{4}{7} \]

### Step 11: Calculating the Percent Degree of Association
The percent degree of association is given by:

\[ \text{Percent association} = \alpha \times 100 = \frac{4}{7} \times 100 \approx 57.14\% \]

### Final Answer
Thus, the percent degree of association of oxygen is approximately **60%**.

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