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

##### 1 Answer

If you look at your units, it becomes easier.

Let us define

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 **per-mol basis**, so you should either convert the mass of the water to **per-gram basis**.

I'll be converting the mass of the water to

#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"#