Question #cf988

Yes, you are correct.

The molar enthalpy of vaporization, or simply the enthalpy of vaporization, tells you the enthalpy change that occurs when one mole of a substance goes from liquid at its boiling point to vapor at its boiling point.

Since heat is needed in order for a substance to undergo a liquid #-># vapor phase change, the enthalpy of vaporization, #DeltaH_"vap"#, will carry a positive sign.

In your case, the value

#DeltaH_"vap" = + "39.23 kJ/mol"#

tells you that #"39.23 kJ"# of heat are being given off when one mole of methanol undergoes a liquid #-># vapor phase change.

Convert the mass of methanol to moles by using the compound's molar mass

#10.0 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"OH")/(32.04color(red)(cancel(color(black)("g")))) = "0.3121 moles CH"_3"OH"#

So, if one mole gives off #"39.23 kJ"# of heat, it follows that #0.3121# moles will give off

#0.3121 color(red)(cancel(color(black)("moles CH"_3"OH"))) * "39.23 kJ"/(1color(red)(cancel(color(black)("mole CH"_3"OH")))) = "12.24 kJ"#

The enthalpy change will thus be

#DeltaH = color(green)(+"12.2 kJ")#

The answer is rounded to three sig figs.