Question

How much heat in kilocalories is required to melt 2.0 mol of isopropyl alcohol (rubbing alcohol; molar mass = 60.0 g/mol)? The heat of fusion and heat of vaporization of isopropyl alcohol are 21.4 cal/g and 159 cal/g, respectively. Pt. B: How much heat in kilocalories is required to boil 2.0 mol ??

Answer 1

First convert moles to mass in grams by multiplying with molar mass of isopropyl alcohol

2.0 mol x 60.0 g/mol = 120 g

Part-A: Melting:

Q = mass in grams x heat of fusion = 120 g x 21.4 cal/g = 2568 cal = 2.568 kcal

Answer: 2.568 kcal of heat is required to melt 2.0 mol of isopropyl alcohol.

Part-B: Boiling:

Q = mass in g x heat of vaporization = 120 g x 159 cal/g = 19080 cal = 19.08 kcal

Answer: 19.08 kcal of heat is required to boil 2.0 mol of isopropyl alcohol.

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