Question

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

Answer 1

If you look at your units, it becomes easier.

Let us define `DeltabarH_"vap"` as the molar enthalpy of vaporization in `"kJ/mol"`, and `DeltaH_"vap"` as the enthalpy of vaporization in `"kJ"`.

A simple equation relating them is:

`\mathbf(DeltaH_"vap" = n_("H"_2"O")DeltabarH_"vap")`

where `n_("H"_2"O")` is just the number of `"mol"`s of water.

You have a mass in `"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 `"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 `"mol"`s of water.

`n_("H"_2"O") = 63.5` `cancel("g H"_2"O") xx ("1 mol H"_2"O")/(18.015 cancel("g H"_2"O"))`

`= color(green)("3.525 mol H"_2"O")`

Hence, we just have:

`DeltaH_"vap" = n_("H"_2"O")DeltabarH_"vap"`

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

`=` `color(blue)"143.3 kJ"`