# How do you calculate enthalpy of vaporization for 1 mole of water?

Aug 8, 2018

The enthalpy of vaporization is $+ 40 , 680 J \mathmr{and} + 40.68 K J$

#### Explanation:

Enthalpy or Heat of vaporization is given by;

$Q = m {H}_{v}$

Where;

${H}_{v} = 2260 J {g}^{-} 1 \left(\text{water at} {100}^{o} C\right)$

$Q = \text{Enthalpy or Heat of vaporization}$

$m = \text{mass}$

Now we are given $1$ mole of water..

Recall;

$\text{No of moles" = "mass"/"molar mass}$

$n = \frac{m}{M m}$

$\therefore m = n \times M m$

$n = 1 \text{moles}$

$M m \textcolor{w h i t e}{x} o f \textcolor{w h i t e}{x} {H}_{2} O = \left(1 \times 2\right) + 16 = 2 + 16 = 16 g m o {l}^{-} 1$

Hence;

$m = 1 \cancel{m o l} \times 18 g \cancel{m o {l}^{-} 1}$

$m = 18 g$

Now plugging it into the formula;

$Q = 18 g \times 2260 J {g}^{-} 1$

$Q = 18 \cancel{g} \times 2260 J \cancel{{g}^{-} 1}$

$Q = 40 , 680 J \mathmr{and} 40.68 K J$

Therefore the enthalpy of vaporization is $+ 40 , 680 J \mathmr{and} + 40.68 K J$