Calculate the enthalpy of vaporisation per mole for ethanol. Given `DeltaS = 109.8 J K^(-1) mol^(-1)` and boiling point of ethanol is `78.5^(@)`.

Explanation

We know that,
`DeltaS_("vapour")=(DeltaH_("Vapour"))/(T_(bp))`
Given, `DeltaS_("vapour")=109.8JK^(-1)mol^(-1)`
`T_(bp)=78.5+273=351.5K`
Substituting these values in above equation, we get
`109.8=(DeltaH_("vapour"))/(351.5)`
`DeltaH_("vapour")=38594.7J" "mol^(-1)`
`=38.594kJ" "mol^(-1)`.