When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are sumulataneously inserted into opposite ends of a glass tube 87.0 cm long, a white ring of `NH_(4)Cl` forms where gaseous `NH_(3)` and gaseous HCl first come into contact.
`NH_(3)(g)+HCl(g)rarrNH_(4)Cl(s)`
At what distance from the ammonis-moistened plug does this occur?

Options

A. 51.7 cm from `NH_(3)` end
B. 51.7 cm from HCl end
C. 43.5 at mid point
D. None of these

Explanation

To solve the problem, we need to determine the distance from the ammonia-moistened plug to the point where the white ring of ammonium chloride (NH₄Cl) forms in a glass tube that is 87.0 cm long.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a glass tube that is 87.0 cm long.
- One end of the tube has a cotton plug moistened with ammonia (NH₃) and the other end has a cotton plug moistened with hydrochloric acid (HCl).

2. **Defining Variables**:
- Let \( x \) be the distance from the ammonia-moistened plug to the point where the white ring forms.
- The distance from the hydrochloric acid plug to the white ring will then be \( 87.0 - x \).

3. **Using 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.
- The formula can be expressed as:
\[ \frac{R_{NH_3}}{R_{HCl}} = \frac{\sqrt{M_{HCl}}}{\sqrt{M_{NH_3}}} \]
- Where \( R \) is the rate of diffusion and \( M \) is the molar mass.

4. **Calculating Molar Masses**:
- Molar mass of NH₃ (ammonia) = 17.0 g/mol
- Molar mass of HCl (hydrochloric acid) = 36.5 g/mol

5. **Setting Up the Equation**:
- The ratio of the distances traveled by the gases is equal to the inverse ratio of their rates of diffusion:
\[ \frac{x}{87.0 - x} = \frac{\sqrt{36.5}}{\sqrt{17.0}} \]

6. **Calculating the Square Roots**:
- Calculate the square roots:
\[ \sqrt{36.5} \approx 6.04 \quad \text{and} \quad \sqrt{17.0} \approx 4.12 \]
- Therefore, the ratio becomes:
\[ \frac{x}{87.0 - x} = \frac{6.04}{4.12} \]

7. **Cross-Multiplying to Solve for \( x \)**:
- Cross-multiply:
\[ 4.12x = 6.04(87.0 - x) \]
- Expanding the right side:
\[ 4.12x = 525.48 - 6.04x \]
- Combine like terms:
\[ 4.12x + 6.04x = 525.48 \]
\[ 10.16x = 525.48 \]
- Solve for \( x \):
\[ x = \frac{525.48}{10.16} \approx 51.7 \text{ cm} \]

8. **Conclusion**:
- The distance from the ammonia-moistened plug to the point where the white ring forms is approximately **51.7 cm**.