Question

The atomic masses of \( ^1H = 1.673 \times 10^{-27} \, \text{kg} \) and \( ^{35}Cl = 58.06 \times 10^{-27} \, \text{kg} \). The reduced mass of HCl is:

• A. \(162.6 \times 10^{-27} \, \text{kg}\)
• B. \(16.26 \times 10^{-27} \, \text{kg}\)
• C. \(1.626 \times 10^{-27} \, \text{kg}\)
• D. \(1626 \times 10^{-27} \, \text{kg}\)

Hint:

The reduced mass is crucial for calculations involving molecular vibrations and rotational spectra.

Answer 1

The Correct Option is C

The reduced mass (\( \mu \)) of the HCl molecule is calculated using the formula: \[ \mu = \frac{m_1 m_2}{m_1 + m_2} \] where:

• \( m_1 \) (mass of \( ^1\text{H} \)): \( 1.673 \times 10^{-27} \, \text{kg} \)
• \( m_2 \) (mass of \( ^{35}\text{Cl} \)): \( 58.06 \times 10^{-27} \, \text{kg} \)

1. Step 1: Substitute Values
   \[ \mu = \frac{(1.673 \times 10^{-27} \, \text{kg}) \times (58.06 \times 10^{-27} \, \text{kg})}{1.673 \times 10^{-27} \, \text{kg} + 58.06 \times 10^{-27} \, \text{kg}} \]
2. Step 2: Calculate Numerator and Denominator
   \[ \text{Numerator} = (1.673 \times 58.06) \times 10^{-54} \, \text{kg}^2 = 97.12 \times 10^{-54} \, \text{kg}^2 \] \[ \text{Denominator} = (1.673 + 58.06) \times 10^{-27} \, \text{kg} = 59.733 \times 10^{-27} \, \text{kg} \]
3. Step 3: Calculate Reduced Mass
   \[ \mu = \frac{97.12 \times 10^{-54} \, \text{kg}^2}{59.733 \times 10^{-27} \, \text{kg}} = 1.627 \times 10^{-27} \, \text{kg} \]
4. Step 4: Round for Significant Figures
   With masses given to three significant figures: \[ \mu \approx 1.626 \times 10^{-27} \, \text{kg} \]

Conclusion:
The reduced mass of HCl is approximately \( 1.626 \times 10^{-27} \, \text{kg} \). The correct answer is: \[ \boxed{(1) \, 1.626 \times 10^{-27} \, \text{kg}} \] Note:
There seems to be a typo in the original options. Option (3) shows \( 1626 \times 10^{-27} \, \text{kg} \), which is significantly off. The calculated reduced mass matches option (1) or (4), both being \( 1.626 \times 10^{-27} \, \text{kg} \).