What is the thermal energy needed to vaporize 18.02g of water at 100°C? The Heat of Vaporization for water is 40.650 kJ/mol.

40.7 kJ

Explanation:

Applying,

q = c'n Equation 1

Where q = Thermal Heat, c' = Heat of vaporization of water, n = number of mole of water.

But,

n = mass(m)/Molar mass(m')

n = m/m' Equation 2

Substitute equation 2 into equation 1

q = c'(m/m') Equation 3

Given: c' = 40.650 KJ/mol, m = 18.02 g

Constant: m' = 18 g/mol

Substitute into equation 3

q = 40.650(18.02/18)

q = 40.695 kJ

q ≈ 40.7 kJ

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