Final answer:
The heat of vaporization for ethanol is 0.826 kJ/g. When 2.19 g of water boils at atmospheric pressure, approximately 1805 J of heat is absorbed.
Step-by-step explanation:
The heat of vaporization for ethanol is 0.826 kJ/g. To calculate the amount of heat absorbed when 2.19 g of water boils at atmospheric pressure, we can use the formula Q = mLv, where Q is the heat absorbed, m is the mass of the substance, and Lv is the heat of vaporization.
First, we need to convert the heat of vaporization from kJ/g to J/g by multiplying it by 1000. So, the heat of vaporization for ethanol is 826 J/g.
Using the formula, Q = (2.19 g)(826 J/g) = 1804.94 J (to 3 significant figures). Therefore, approximately 1805 J of heat is absorbed when 2.19 g of water boils at atmospheric pressure.