Question

UnansweredYou AnsweredCorrect AnswerQuestion 90/1 ptsHow much energy (kJ) is required to vaporize 92.3 g ofCH₂Cl2 at its boiling point, if its AHvap is 29.8 kJ/mol?Enter your answer to 1 decimal place.32.4 margin of error +/-0.1

Answer 1

To answer the question we will use the following equation:

`Q=\Delta H_(vap).n`

Where:

Q is the energy we need to vaporize the substance

Hvap is he enthalpy of vaporization

n is the number of moles

We first calculate the number of moles by using the molar mass of CH2Cl2:

`\begin{gathered} M_(CH_2Cl_2)=M_C+2M_H+2M_(Cl) \\ M_(CH_2Cl_2)=12(g)/(mol)+2.1(g)/(mol)+2.35.45(g)/(mol)=84.9(g)/(mol) \end{gathered}`

We calculate n (number of moles):

`n=(92.3g)/(84.9(g)/(mol))=1.087mol`

Now we calculate Q (energy)

`Q=29.8(kJ)/(mol).1.087mol=32.4kJ`

So the answer is 32.4kJ