Question

Calculate the moment of inertia of a CH³⁵CL₃ molecule around a rotational axis that contains the C-H bond. The C-Cl bond length is 177pm and the HCCl angle is 107⁰f​

Answer 1

Answer:

The correct answer is "$4.991\times 10^{-45} \ kg.m^2$".

Explanation:

According to the question,

$R_{C-Cl} = 177 \ pm$

or,

         $=1.77\times 10^{-10} \ m$

$\alpha = 107^{\circ}$

$m_{Cl}=34.97 \ m.u$

or,

      $=34.97\times 1.66\times 10^{-27}$

      $=5.807\times 10^{-26} \ kg$

The moment of inertia around the rotational axis will be:

⇒  $I=3\times m_{Cl}\times (R_{C-Cl})^2 \ Sin^2 \alpha$

By putting the values, we get

       $=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2 \ Sin^2 (107)$

       $=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2\times 0.91452$

       $=4.991\times 10^{-45} \ kg.m^2$