Science & Mathematics » Chemistry » The heat of vaporization of water is 40.66 kj/mole. How much heat is absorbed when 2.74 g of water boils at atmospheric pressure?

The heat of vaporization of water is 40.66 kj/mole. How much heat is absorbed when 2.74 g of water boils at atmospheric pressure?

1 mole (18g) of water releases 40.66kJ. Then 2.74g (0.1522 mole) of water would release 0.1522 mole x 40.66kJ/mole = 6.19kJ

40.66 kj/mole = 40,660 J/mole
The mass of one mole of water is 18 grams.
Number of moles = 2.74 ÷ 18
Q = (2.74 ÷ 18) * 40,660
This is approximately 6189 J