The heat of vaporization of ammonia (NH3) is 327 cal/g. What is the molar heat of vaporization of ammonia in kilojoules per mole?
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Use dimensional analysis:
1 cal = 4.186 J
1 kJ = 1,000 J
1 mol = 17.03052 g
[(327 cal)(1 g)][(4.186 J)/(1 cal)][(1 kJ)/(1,000 J)][(17.03052 g)/(1 mol)] = 23.31175 kJ/mol or 23.3 kJ/mol rounded to three significant figures
Answer: The molar heat of vaporization of ammonia is about 23.3 kJ/mol.
1 cal = 4.186 J
1 kJ = 1,000 J
1 mol = 17.03052 g
[(327 cal)(1 g)][(4.186 J)/(1 cal)][(1 kJ)/(1,000 J)][(17.03052 g)/(1 mol)] = 23.31175 kJ/mol or 23.3 kJ/mol rounded to three significant figures
Answer: The molar heat of vaporization of ammonia is about 23.3 kJ/mol.
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Molar mass of NH3 is 17.0306 g/mol
17.03g/m*327cal/g=5559cal or 5560 cal/m (sig figs) =5.56 Kcal/m
5.56 Kcal/m *4.184Kj/Kcal =23.3Kj/m
17.03g/m*327cal/g=5559cal or 5560 cal/m (sig figs) =5.56 Kcal/m
5.56 Kcal/m *4.184Kj/Kcal =23.3Kj/m