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Answer :
The amount of heat required to melt 1.70 mol of isopropyl alcohol (rubbing alcohol) is 7,224 kilocalories. Heat in kilocalories is required to melt 1.70 mol of isopropyl alcohol
This is calculated by using the molar mass of the alcohol (60.10 g/mol) multiplied by the heat of fusion (4.2 kcal/mol). The equation looks like this: (1.70 mol) x (60.10 g/mol) x (4.2 kcal/mol) = 7,224 kcal.
The heat of fusion is the amount of energy needed to convert one mole of a substance from a solid phase to a liquid phase.
This occurs at the melting point of the particular material. In the case of isopropyl alcohol, the melting point is -89.5°C. Thus, at this temperature, the molecules of the substance have enough energy to break apart from the solid lattice structure and become liquid. The energy needed to do this is the heat of fusion.
Heat in kilocalories is required to melt 1.70 mol of isopropyl alcohol.
Know more about Molar mass here.
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