The reaction between gaseous `NH_3` and HBr produces a white solid `NH_4Br`. Suppose that `NH_3` and HBr are introduced simultaneously into the opposite ends of an open tube of 1 metre length. Where would you expect the white solid to form?

Options

A. At a distance of 34.45 cm from `NH_3` end
B. At a distance of 68.5 cm from `NH_3` end
C. At a distance of 44.45 cm from HBr end
D. At a distance of 45.45 cm from HBr end

Answer

To solve the problem of where the white solid ammonium bromide (NH₄Br) will form when gaseous ammonia (NH₃) and hydrogen bromide (HBr) are introduced simultaneously into opposite ends of a 1-meter long tube, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- The reaction between NH₃ and HBr produces NH₄Br, which is a white solid. We need to determine where this solid will form in the tube.

2. **Setting Up the Problem**:
- Let’s denote the distance traveled by NH₃ as \( l \) cm from its end, and the distance traveled by HBr as \( 100 - l \) cm from its end (since the total length of the tube is 100 cm).

3. **Applying Graham's Law of Effusion**:
- According to Graham's law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Therefore, we can write:
\[ \frac{\text{Rate of diffusion of NH₃}}{\text{Rate of diffusion of HBr}} = \frac{\sqrt{M_{\text{HBr}}}}{\sqrt{M_{\text{NH₃}}}} \]
- Where \( M_{\text{HBr}} \) is the molar mass of HBr and \( M_{\text{NH₃}} \) is the molar mass of NH₃.

4. **Calculating Molar Masses**:
- Molar mass of NH₃ (Nitrogen = 14, Hydrogen = 3):
\[ M_{\text{NH₃}} = 14 + (3 \times 1) = 17 \, \text{g/mol} \]
- Molar mass of HBr (Hydrogen = 1, Bromine = 80):
\[ M_{\text{HBr}} = 1 + 80 = 81 \, \text{g/mol} \]

5. **Setting Up the Equation**:
- Using Graham's law, we can set up the equation:
\[ \frac{l}{100 - l} = \frac{\sqrt{81}}{\sqrt{17}} \]
- This simplifies to:
\[ \frac{l}{100 - l} = \frac{9}{\sqrt{17}} \]

6. **Cross-Multiplying and Solving for \( l \)**:
- Cross-multiplying gives:
\[ l \cdot \sqrt{17} = 9(100 - l) \]
- Expanding and rearranging:
\[ l \cdot \sqrt{17} + 9l = 900 \]
\[ l(\sqrt{17} + 9) = 900 \]
\[ l = \frac{900}{\sqrt{17} + 9} \]

7. **Calculating the Value of \( l \)**:
- Approximating \( \sqrt{17} \approx 4.123 \):
\[ l \approx \frac{900}{4.123 + 9} \approx \frac{900}{13.123} \approx 68.5 \, \text{cm} \]

8. **Conclusion**:
- Therefore, the white solid NH₄Br will form at a distance of approximately **68.5 cm from the NH₃ end** of the tube.