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how much heat in kilocalories is required to melt 8.65 moles 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. Round your answer to the tenths of a kcal without units.

User AaronS
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Final answer:

To melt 8.65 moles of isopropyl alcohol, 11.1 kilocalories of heat are required. This is calculated by converting moles to grams, multiplying by the heat of fusion, and converting calories to kilocalories.

Step-by-step explanation:

To calculate the amount of heat in kilocalories required to melt 8.65 moles of isopropyl alcohol, we first need to find the total heat in calories needed by using the heat of fusion value and then convert the total heat to kilocalories. The heat of fusion of isopropyl alcohol is 21.4 cal/g. To begin, we convert moles to grams using the molar mass of isopropyl alcohol which is 60.0 g/mol.

8.65 moles of isopropyl alcohol × 60.0 g/mol = 519 g of isopropyl alcohol

Next, we multiply the mass of isopropyl alcohol by its heat of fusion:

519 g × 21.4 cal/g = 11106.6 calories

Now, to convert calories to kilocalories, we divide by 1000, as there are 1000 calories in a kilocalorie:

11106.6 calories ÷ 1000 = 11.1 kilocalories

Therefore, 11.1 kilocalories of heat are required to melt 8.65 moles of isopropyl alcohol.

User Chrislusf
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