Question#19308
How much heat in kilocalories is required to melt 1.0mole of isopropyl alcohol (rubbing alcohol; molar mass = 60.0g/mol )? The heat of fusion and heat of vaporization of isopropyl alcohol are 21.4cal/g and 159cal/g , respectively
Solution:
Q=cm, were m – mass, c - Enthalpy of fusion (heat of fusion),
Such as 1 mole=60 g, m=60 g,
Q=21.4*60=1284 cal = 1.284 Kcal
Answer: 1.284 Kcal.