`O_(2)` is partially atomised due to certain experimental conditions. The mixture of ` O_(2)` molecules and O atoms diffuses `sqrt(5)` times slower than Helium. What is the percentage atomisation of `O_(2)` ?

Answer

To solve the problem of determining the percentage atomization of \( O_2 \), we can follow these steps:

### Step 1: Understand the relationship between diffusion rates and molecular masses.
The rate of diffusion of gases is inversely proportional to the square root of their molecular masses. This means that if we denote the rate of diffusion of helium as \( R_{He} \) and the rate of diffusion of the mixture of \( O_2 \) and \( O \) as \( R_{mix} \), we can express this relationship as:
\[ \frac{R_{He}}{R_{mix}} = \sqrt{5} \]

### Step 2: Relate the diffusion rates to molecular masses.
From the relationship above, we can also write:
\[ \frac{M_{mix}}{M_{He}} = 5 \]
where \( M_{mix} \) is the molecular mass of the mixture and \( M_{He} \) is the molecular mass of helium (which is 4 g/mol). Thus, we have:
\[ M_{mix} = 5 \times M_{He} = 5 \times 4 = 20 \text{ g/mol} \]

### Step 3: Set up the equation for the average molecular mass of the mixture.
Let \( X \) be the percentage of \( O_2 \) in the mixture. Then the percentage of \( O \) (oxygen atoms) will be \( 100 - X \). The molecular masses are:
- \( O_2 \): 32 g/mol
- \( O \): 16 g/mol

The average molecular mass \( M_{mix} \) can be expressed as:
\[ M_{mix} = \frac{X \cdot 32 + (100 - X) \cdot 16}{100} \]

### Step 4: Set the average molecular mass equal to 20 g/mol.
Now we can set up the equation:
\[ \frac{X \cdot 32 + (100 - X) \cdot 16}{100} = 20 \]

### Step 5: Solve for \( X \).
Multiplying both sides by 100 gives:
\[ X \cdot 32 + (100 - X) \cdot 16 = 2000 \]
Expanding this:
\[ 32X + 1600 - 16X = 2000 \]
Combining like terms:
\[ 16X + 1600 = 2000 \]
Subtracting 1600 from both sides:
\[ 16X = 400 \]
Dividing by 16:
\[ X = 25 \]

### Step 6: Calculate the percentage atomization of \( O_2 \).
The percentage of \( O_2 \) in the mixture is 25%. Therefore, the percentage of atomization of \( O_2 \) is:
\[ 100 - X = 100 - 25 = 75\% \]

### Final Answer:
The percentage atomization of \( O_2 \) is **75%**.

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