# Enthalpy of Vaporization

## Practice Problems

### Problem 1

The enthalpy of vaporization of water is ΔHvap = 44.01 kJ mol–1 at standard temperature. How much heat (in kJ) is required to vaporize exactly two moles of water at that temperature?

Solution

\begin{align} 2~\mathrm{mol} \left ( \dfrac{44.01~\mathrm{kJ}}{\mathrm{mol}} \right ) = 88.02~\mathrm{kJ} \end{align}

## Timed Assessments

### Heat of Vaporization - q

VIDEO

1. 1.13 × 103 kJ
2. 6.09 × 102 kJ
3. 2.16 × 102 kJ

VIDEO

1. 1.80 L
2. 1.60 L
3. 1.56 L

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