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

The correct answer is "
`4.991* 10^(-45) \ kg.m^2`".

Step-by-step explanation:

According to the question,

`R_(C-Cl) = 177 \ pm`

or,

`=1.77* 10^(-10) \ m`

`\alpha = 107^(\circ)`

`m_(Cl)=34.97 \ m.u`

or,

`=34.97* 1.66* 10^(-27)`

`=5.807* 10^(-26) \ kg`

The moment of inertia around the rotational axis will be:

⇒
`I=3* m_(Cl)* (R_(C-Cl))^2 \ Sin^2 \alpha`

By putting the values, we get

`=3* 5.807* 10^(-26)* (1.77* 10^(-10))^2 \ Sin^2 (107)`

`=3* 5.807* 10^(-26)* (1.77* 10^(-10))^2* 0.91452`

`=4.991* 10^(-45) \ kg.m^2`