MasteringChemistry: Post-Lecture Homework Chapter 11 - Microsoft Edge https://se

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MasteringChemistry: Post-Lecture Homework Chapter 11 - Microsoft Edge https://session.masteringchemistry.com/myct/itemVi mID-96617028 Post-Lecture Homework Chapter 11 Problem 11.46 14 of 22 Part A n compound C2 Cla Fa has a normal boiling The f point of 47.6 C. The specific heats of C2Cl3F3() and C2 ClaF3 (9 are 0.91 J/g K and 0.67 J/gK respectively. The heat of vaporization for the compound is 27.49 kJ/mol. Calculate the heat required to convert 72.5 g of C2 ClFs from a liquid at 10.40 °C to a gas at 82.45 °C. Express your answer using two significant figures. k.J Submit Provide Feedback Next > 1:17 AM 0 Type here to search 1/14/2018

Explanation / Answer

Ti = 10.4 oC
Tf = 82.45 oC
here
Cl = 0.91 J/g.K = 0.91 J/g.oC

Heat required to convert liquid from 10.4 oC to 47.6 oC
Q1 = m*Cl*(Tf-Ti)
= 16.5 g * 0.91 J/g.oC *(47.6-10.4) oC
= 558.558 J

Hvap = 27.49KJ/mol = 27490J/mol


Lets convert mass to mol
Molar mass of C2Cl3F3 = 187.37 g/mol
number of mol
n= mass/molar mass
= 16.5/187.37
= 0.0881 mol

Heat required to convert liquid to gas at 47.6 oC
Q2 = n*Hvap
= 0.0881 mol *27490 J/mol
= 2420.7984 J

Cg = 0.67 J/g.K = 0.67 J/g.oC

Heat required to convert vapour from 47.6 oC to 82.45 oC
Q3 = m*Cg*(Tf-Ti)
= 16.5 g * 0.67 J/g.oC *(82.45-47.6) oC
= 385.2668 J

Total heat required = Q1 + Q2 + Q3
= 558.558 J + 2420.7984 J + 385.2668 J
= 3364.6232 J
= 3.4 KJ
Answer: 3.4 KJ

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