A gaseous mixture of `O_(2)` and an unknown gas 'X' containing 20 mole `%` of X diffused through a small hole in 245 seconds while `O_(2)` takes 220 seconds to diffuse through the same hole under similar conditions. Calculate the molecular mass of X.

Explanation

To solve the problem, we will use Graham's law of diffusion, which relates the rates of diffusion of two gases to their molar masses. Here’s a step-by-step solution:

### Step 1: Understand the given data
- The time taken for gas X to diffuse (T_X) = 245 seconds
- The time taken for O₂ to diffuse (T_O₂) = 220 seconds
- The mole percentage of gas X = 20%
- The mole percentage of O₂ = 80%
- The molar mass of O₂ (M_O₂) = 32 g/mol

### Step 2: Apply Graham's law of diffusion
According to Graham's law:
\[ \frac{R_X}{R_{O_2}} = \sqrt{\frac{M_{O_2}}{M_X}} \]
Where \( R \) is the rate of diffusion, which can be expressed as:
\[ R = \frac{\text{Amount diffused}}{\text{Time taken}} \]
Since the amount diffused is constant, we can express the rates in terms of time:
\[ \frac{T_{O_2}}{T_X} = \sqrt{\frac{M_X}{M_{O_2}}} \]

### Step 3: Substitute the known values
Substituting the known values into the equation:
\[ \frac{220}{245} = \sqrt{\frac{M_X}{32}} \]

### Step 4: Square both sides to eliminate the square root
\[ \left(\frac{220}{245}\right)^2 = \frac{M_X}{32} \]

### Step 5: Calculate the left side
Calculating \( \left(\frac{220}{245}\right)^2 \):
\[ \frac{220^2}{245^2} = \frac{48400}{60025} \approx 0.8065 \]

### Step 6: Solve for \( M_X \)
Now, substituting back:
\[ 0.8065 = \frac{M_X}{32} \]
Multiplying both sides by 32:
\[ M_X = 0.8065 \times 32 \approx 25.8 \text{ g/mol} \]

### Step 7: Calculate the average molar mass of the mixture
Since gas X constitutes 20% of the mixture, we can calculate the average molar mass of the mixture (M_mixture):
\[ M_{mixture} = 0.8 \times M_{O_2} + 0.2 \times M_X \]
Substituting the values:
\[ M_{mixture} = 0.8 \times 32 + 0.2 \times M_X \]
We already found \( M_X \) to be approximately 25.8 g/mol. Now substituting:
\[ M_{mixture} = 0.8 \times 32 + 0.2 \times 25.8 \]
Calculating:
\[ M_{mixture} = 25.6 + 5.16 = 30.76 \text{ g/mol} \]

### Step 8: Finalize the molecular mass of gas X
Using the average molar mass of the mixture and the mole percentages, we can solve for \( M_X \) again if necessary, but we have already derived it through the diffusion rates.

### Conclusion
The molecular mass of gas X is approximately 25.8 g/mol.

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