1) How much heat energy in kJ must be removed from benzene, C6H6, at its freezin

Question

1) How much heat energy in kJ must be removed from benzene, C6H6, at its freezing point to freeze 75.1 g benzene? The heat of fusion of benzene is 9.95 kJ/mol.

2)Calculate the heat required to convert 65.0g of C2Cl3F3 from a liquid at 10.00?C to a gas at 90.75?C.

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/g?K and 0.67 J/g?K, respectively. The heat of vaporization for the compound is 27.49 kJ/mol.

3) How many grams of water can be cooled from 44?C to 16?C by the evaporation of 45g of water? (The heat of vaporization of water in this temperature range is 2.4 kJ/g. The specific heat of water is 4.18 J/g?K.)

Explanation / Answer

Answer : As we know that molar mass of Benzene = 12* 6 + 6*1 = 78 g/mol

Hence the number of moles of 75.1 g of benzene = 75.1 / 78 = 0.9628 mol

Hence amount of heat needed to freez 75.1 g of benzene = 0.9628 * [-9.95 kj/mol ] = -9.57986 Kj

as weknow heat lose = heat gain

Hence the heat energy revome = 9.57986 Kj

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