What is the enthalpy required to vaporize "63.5 g" of water at 100^@ "C"? DeltabarH_"vap" = "40.66 kJ/mol".

Jan 2, 2016

If you look at your units, it becomes easier.

Let us define $\Delta {\overline{H}}_{\text{vap}}$ as the molar enthalpy of vaporization in $\text{kJ/mol}$, and $\Delta {H}_{\text{vap}}$ as the enthalpy of vaporization in $\text{kJ}$.

A simple equation relating them is:

$\setminus m a t h b f \left(\Delta {H}_{\text{vap" = n_("H"_2"O")DeltabarH_"vap}}\right)$

where ${n}_{\text{H"_2"O}}$ is just the number of $\text{mol}$s of water.

You have a mass in $\text{g}$, and you have a molar enthalpy that is on a per-mol basis, so you should either convert the mass of the water to $\text{mol}$s of water or convert the molar enthalpy to be on a per-gram basis.

I'll be converting the mass of the water to $\text{mol}$s of water.

${n}_{\text{H"_2"O}} = 63.5$ cancel("g H"_2"O") xx ("1 mol H"_2"O")/(18.015 cancel("g H"_2"O"))

$= \textcolor{g r e e n}{\text{3.525 mol H"_2"O}}$

Hence, we just have:

$\Delta {H}_{\text{vap" = n_("H"_2"O")DeltabarH_"vap}}$

= ("3.525" cancel("mol H"_2"O"))("40.66 kJ/"cancel("mol H"_2"O"))

$=$ $\textcolor{b l u e}{\text{143.3 kJ}}$