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. 7
B. 10
C. 9
D. 8

Answer

To solve the problem of determining at which row the students will start laughing and weeping simultaneously, we can follow these steps:

### Step 1: Understand the Setup
We have a classroom with 13 equidistant rows of benches. The teacher releases laughing gas (N₂O) from the first bench (row 1), and the student releases weeping gas (C₆H₁₁OBr) from the last bench (row 13). We need to find the row where both gases meet.

### Step 2: Define the Distances
Let:
- \( x \) = distance traveled by the laughing gas (N₂O) from the first bench.
- \( 13 - x \) = distance traveled by the weeping gas (C₆H₁₁OBr) from the last bench.

### Step 3: Use Graham's Law of Effusion
According to Graham's law, the rates of effusion of two gases are inversely proportional to the square roots of their molar masses:
\[ \frac{\text{Rate of } N_2O}{\text{Rate of } C_6H_{11}OBr} = \sqrt{\frac{M_{C_6H_{11}OBr}}{M_{N_2O}}} \]

### Step 4: Calculate Molar Masses
1. **Molar mass of C₆H₁₁OBr**:
- Carbon (C): 6 × 12 = 72 g/mol
- Hydrogen (H): 11 × 1 = 11 g/mol
- Oxygen (O): 1 × 16 = 16 g/mol
- Bromine (Br): 1 × 80 = 80 g/mol
- Total = 72 + 11 + 16 + 80 = 179 g/mol

2. **Molar mass of N₂O**:
- Nitrogen (N): 2 × 14 = 28 g/mol
- Oxygen (O): 1 × 16 = 16 g/mol
- Total = 28 + 16 = 44 g/mol

### Step 5: Set Up the Ratio
Using the molar masses calculated:
\[ \frac{\text{Rate of } N_2O}{\text{Rate of } C_6H_{11}OBr} = \sqrt{\frac{179}{44}} \approx \sqrt{4.07} \approx 2 \]

### Step 6: Relate Distances and Rates
Since the time taken for both gases to travel their respective distances is the same, we can set up the equation:
\[ \frac{x}{13 - x} = 2 \]

### Step 7: Solve for \( x \)
Cross-multiplying gives:
\[ x = 2(13 - x) \]
\[ x = 26 - 2x \]
\[ 3x = 26 \]
\[ x = \frac{26}{3} \approx 8.67 \]

### Step 8: Determine the Row
Since \( x \) represents the distance from the first bench, we round \( 8.67 \) to the nearest whole number, which gives us approximately row 9.

### Conclusion
The students will start laughing and weeping simultaneously at approximately **row 9**.

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