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

Answer:

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:

$\begin{gathered} M_{CH_2Cl_2}=M_C+2M_H+2M_{Cl} \\ M_{CH_2Cl_2}=12\frac{g}{mol}+2.1\frac{g}{mol}+2.35.45\frac{g}{mol}=84.9\frac{g}{mol} \end{gathered}$

We calculate n (number of moles):

$n=\frac{92.3g}{84.9\frac{g}{mol}}=1.087mol$

Now we calculate Q (energy)

$Q=29.8\frac{kJ}{mol}.1.087mol=32.4kJ$

So the answer is 32.4kJ