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?

59.87 kJ

Explanation:

In the question, we are given;

Molar heat of Vaporization as 10.79 kJ/mol Mass of water at 100°C is 100 g

We are required to calculate the amount of heat energy absorbed to convert water at 100°C to steam at 100°C.

The process of converting water from liquid state to solid state without change in temperature is known as condensation. We are given the molar heat of vaporization as 10.79 kJ/mol, this means 1 mole of water will absorb 10.79 kJ when converted to ice without change in temperature. Therefore, we can first calculate the number of moles of water;

Moles = Mass : Molar mass

Molar mass of water = 18.02 g/mol

Moles of water = 100 g : 18.02 g/mol

= 5.549 moles

But;

Q = n * ΔHv, where n is the number of moles and ΔHv molar heat of Vaporization.

Therefore;

Q = 5.549 moles * 10.79 kJ/mol

= 59.874 kJ

= 59.87 kJ

Thus, the amount of heat absorbed is 59.87 kJ