# The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 2.87 g of water boils at atmospheric pressure?

Nov 16, 2016

$\text{6.48 kJ}$

#### Explanation:

The molar heat of vaporization, $\Delta {H}_{\text{vap}}$, sometimes called the molar enthalpy of vaporization, tells you how much energy is needed in order to boil $1$ mole of a given substance at its boiling point.

In water's case, a molar heat of vaporization of ${\text{40.66 kJ mol}}^{- 1}$ means that you need to supply $\text{40.66 kJ}$ of heat in order to boil $1$ mole of water at its normal boiling point, i.e. at ${100}^{\circ} \text{C}$.

DeltaH_"vap" = color(blue)("40.66 kJ") color(white)(.)color(red)("mol"^(-1))

You need $\textcolor{b l u e}{\text{40.66 kJ}}$ of heat to boil $\textcolor{red}{\text{1 mole}}$ of water at its normal boiling point.

Now, the first thing to do here is to convert the mass of water to moles by using its molar mass

2.87 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.1593 moles H"_2"O"

You can now use the molar heat of vaporization as a conversion factor to determine how much heat would be needed to boil $0.1593$ moles of water at its boiling point

$0.1593 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles H"_2"O"))) * "40.66 kJ"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("6.48 kJ}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the mass of the sample.