To determine the bond length of an HCl molecule given the moment of inertia (I) and the reduced mass (μ), we follow these steps:
1. Given Information:
- Moment of inertia [tex]\( I = 2.7 \times 10^{-40} \, g \cdot cm^2 \)[/tex]
- Reduced mass [tex]\( \mu = 0.98 \, g \cdot mol^{-1} \)[/tex]
2. Convert the reduced mass from grams per mole to grams per molecule:
- We know that 1 mole of a substance contains Avogadro's number [tex]\((N_A)\)[/tex] of molecules, where [tex]\( N_A = 6.022 \times 10^{23} \)[/tex] molecules per mole.
- Therefore, the reduced mass per molecule is calculated as:
[tex]\[
\mu_{molecule} = \frac{\mu}{N_A}
\][/tex]
[tex]\[
\mu_{molecule} = \frac{0.98 \, g \cdot mol^{-1}}{6.022 \times 10^{23} \, molecules \cdot mol^{-1}}
\][/tex]
3. Calculate the reduced mass per molecule:
[tex]\[
\mu_{molecule} = \frac{0.98}{6.022 \times 10^{23}} \, g
\][/tex]
4. Determine the bond length [tex]\( r \)[/tex] using the relationship between the moment of inertia and the reduced mass:
- The moment of inertia of a diatomic molecule is given by:
[tex]\[
I = \mu_{molecule} \cdot r^2
\][/tex]
- To find the bond length [tex]\( r \)[/tex], we rearrange this formula:
[tex]\[
r = \sqrt{\frac{I}{\mu_{molecule}}}
\][/tex]
5. Substitute the known values into the formula:
[tex]\[
r = \sqrt{\frac{2.7 \times 10^{-40} \, g \cdot cm^2}{\frac{0.98}{6.022 \times 10^{23}} \, g}}
\][/tex]
6. Simplify the expression under the square root:
[tex]\[
r = \sqrt{\frac{2.7 \times 10^{-40} \times 6.022 \times 10^{23}}{0.98}} \, cm
\][/tex]
7. Calculate the numerical value:
[tex]\[
r \approx 1.2880692718093976 \times 10^{-8} \, cm
\][/tex]
Thus, the bond length of the HCl molecule is approximately [tex]\( 1.2880692718093976 \times 10^{-8} \)[/tex] cm.
