A teacher enters a classroom from front door while a student from back door. There are 13 equidistant rows of benches in the classroom. The teacher releases `N_(2)O`, the laughing gas, from the first bench while the student releases the weeping gas `(C_(6)H_(11)OBr)` from the last bench. At which row will the students starts laughing and weeping simultaneously?

Options

A. `9^(th)` row
B. `12^(th)` row
C. `7^(th)` row
D. `10^(th)` row

Explanation

To solve the problem, we need to determine at which row the students will start laughing and weeping simultaneously when the teacher releases laughing gas (N₂O) from the first bench and the student releases weeping gas (C₆H₁₁OBr) from the last bench in a classroom with 13 rows of benches.

### Step-by-Step Solution:

1. **Understand the Concept of Diffusion**:
The rate of diffusion of a gas is inversely proportional to the square root of its molar mass. This can be expressed mathematically as:
\[ R \propto \frac{1}{\sqrt{m}} \]
where \( R \) is the rate of diffusion and \( m \) is the molar mass of the gas.

2. **Define Variables**:
Let:
- \( R_1 \) = Rate of diffusion of N₂O (laughing gas)
- \( R_2 \) = Rate of diffusion of C₆H₁₁OBr (weeping gas)
- \( m_1 \) = Molar mass of N₂O
- \( m_2 \) = Molar mass of C₆H₁₁OBr

3. **Set Up the Relationship**:
From the diffusion principle, we have:
\[ \frac{R_1}{R_2} = \sqrt{\frac{m_2}{m_1}} \]

4. **Calculate Molar Masses**:
- Molar mass of N₂O (laughing gas) = 44 g/mol
- Molar mass of C₆H₁₁OBr (weeping gas) = 179 g/mol

5. **Substitute the Values**:
Substitute the molar masses into the equation:
\[ \frac{R_1}{R_2} = \sqrt{\frac{179}{44}} \]

6. **Calculate the Ratio**:
Calculate the square root:
\[ \sqrt{\frac{179}{44}} \approx 2 \]
Therefore, we have:
\[ \frac{R_1}{R_2} = 2 \]

7. **Define Rows**:
Let \( X \) be the row number from the front (where the teacher is) where the students start laughing and weeping simultaneously. The total number of rows is 13, so:
\[ R_1 = X \quad \text{and} \quad R_2 = 13 - X \]

8. **Set Up the Equation**:
From the ratio we established:
\[ \frac{X}{13 - X} = 2 \]

9. **Cross-Multiply and Solve**:
Cross-multiplying gives:
\[ X = 2(13 - X) \]
Expanding this:
\[ X = 26 - 2X \]
Rearranging gives:
\[ 3X = 26 \implies X = \frac{26}{3} \approx 8.67 \]

10. **Round to Nearest Whole Number**:
Since row numbers must be whole numbers, we round \( 8.67 \) to \( 9 \).

11. **Conclusion**:
Therefore, the students will start laughing and weeping simultaneously at the **9th row**.

### Final Answer:
The students will start laughing and weeping simultaneously at the **9th row**.

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