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The flourocarbon compound C2Cl3F3 has a normal boiling point of 47.6oC. The specific heats of C2Cl3F3(l)...

The flourocarbon compound C2Cl3F3 has a normal boiling point of 47.6oC. The specific heats of C2Cl3F3(l) and C2Cl3F3(g) are 0.91 J/g-K and 0.67 J/g-K, respectively. The heat of vaporization for the compound is 27.49 kJ/mol. Calculate the heat required to convert 50.0 grams of C2Cl3F3 from a liquid at 10 degrees Celcius to a gas at 85 degrees celcius.

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Answer #1

no of moles of C2Cl3F3 = W/G.M.Wt

                                    = 50/187.5   = 0.26 moles

q1 = mc\DeltaT

     = 50*0.91*(47.6-10)

    = 1710.8J   = 7.7108KJ

q2   = n\DeltaHvap

       = 0.26*27.49 = 7.1474KJ

q3    = mc\DeltaT

       = 50*0.67*(85-47.6)   = 1252.9J    = 1.2529KJ

Total heat energy is required = q = q1+q2+q3

                                                   = 7.7108 +7.1474+1.2529   = 16.11KJ

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