The molar heat of vaporization for water is 10.79 kJ/mol. How much energy must be absorbed by 100 grams

of water at 100 °C in order to convert it to steam at 100 °C?

A 4079 0

B 140.79 k)

C 7.35 k)

D 226 kJ

59.878 Kilo joules

Explanation:

We are given;

Molar heat of vaporization for water as 10.79 kJ/mol Mass of water 100 grams

We are required to calculate the amount of energy absorbed to convert water to ice.

We need to know that the process of conversion of water to steam without change in temperature is known as vaporization or boiling. Therefore, heat absorbed is calculated by multiplying the mass of water or the number of moles by heat of vaporization. That is, Q = m * ΔHf When given the number of moles and the molar heat of vaporization, then Q = n * ΔHf

In this case;

Mass of water = 100 g

But, molar mass of water = 18.02 g/mol

Moles of water = 100 g : 18.02 g/mol

= 5.549 Moles

Therefore;

Q = 5.549 moles * 10.79 kJ/mol

= 59.878 Kilo joules