Question

Calculate the amount of heat energy required to evaporate 27.5 g of water at 100.0°C. (molar heat of vaporization of liquid water = 4.07 ✕ 104 J/mol)
a. 6.20 ✕ 107 kJ
b. 2.02 ✕ 104 kJ
c. 1.22 ✕ 10-2 kJ
d. 6.20 ✕ 104 J
e. 1.12 ✕ 103 kJ

Answer 1

Answer:

D. 6.20 × 10^4 J

Explanation:

The heat change in changing water from a liquid state to a gaseous state without change in temperature is given by the formula;

Heat = MLv ; where m is the mass and Lv is the latent heat of vaporization.

In this case we will use mass in terms of number of moles as the latent heat is given in joules/mole.

Therefore;

Moles of water = 27.5 g/18 g/mol

                         = 1.528 moles

Hence;

Heat = 1.528 moles × 4.07 ×10^4 J/mol

        = 6.20 × 10^4 Joules