Problem 11.46 The fluorocarbon compound C2Cl3F3 has a normal boiling point of 47
ID: 981905 • Letter: P
Question
Problem 11.46
The fluorocarbon compound C2Cl3F3 has a normal boiling point of 47.6 C. The specific heats of C2Cl3F3(l) and C2Cl3F3(g) are 0.91 J/gK and 0.67 J/gK, respectively. The heat of vaporization for the compound is 27.49 kJ/mol.
Part A
Calculate the heat required to convert 72.0 g of C2Cl3F3 from a liquid at 10.30 C to a gas at 76.05 C.
Express your answer using two significant figures.
14.83
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Problem 11.46
The fluorocarbon compound C2Cl3F3 has a normal boiling point of 47.6 C. The specific heats of C2Cl3F3(l) and C2Cl3F3(g) are 0.91 J/gK and 0.67 J/gK, respectively. The heat of vaporization for the compound is 27.49 kJ/mol.
Part A
Calculate the heat required to convert 72.0 g of C2Cl3F3 from a liquid at 10.30 C to a gas at 76.05 C.
Express your answer using two significant figures.
Q =14.83
kJSubmitMy AnswersGive Up
Incorrect; Try Again; 6 attempts remaining
Explanation / Answer
Molar mass of C2Cl3F3 = 2 * 12 + 3 * 35.5 + 3 * 19 = 187.5 gm/mol
Number of moles of C2Cl3F3 = 72/187.5 = 0.384 moles
Heat of vaporization for 0.384 moles = 0.384 moles * 27.49 KJ/mol = 10.55616 KJ
Heat required to convert 72g of C2F3Cl3 from the liquid temperature of 10.30C to 47.6C
=> 72 * 0.91 * (47.6-10.3)
=> 2443.896 J
Heat required to convert 72g of C2F3Cl3 from the gaseous temperature of 47.6C to 76.05C
=> 72 * 0.67 * (76.05-47.6)
=> 1372.428 J
Total heat required = 10556.16 J + 1372.428 J + 2443.896 J
=> 14372.424 J = 14.372 KJ