# Molar entropy of vaporization

If a substance has a molar heat of vaporization of $\pu{3.05\times10^4J/mol}$ and a normal boiling temperature of $\pu{80.0^\circ C}$, what is the value of its molar entropy of vaporization $\Delta S_\mathrm{vap}$ in $\pu{J K^{-1} mol^{-1}}$?

I got $$\frac{3.5\times10^4\ \mathrm{J/mol}}{(80+273)~\mathrm K}=86.4~\mathrm{J~K^{-1}~mol^{-1}}$$ however my texbook answer key says that the answer is $380\ \mathrm{J~K^{-1}~mol^{-1}}$. What am I doing wrong?

• It seems like your textbook might have just done 3.05e4/80 = 381, which would be wrong. – a-cyclohexane-molecule Aug 15 '18 at 21:53
• @a-cyclohexane-molecule thank you so much for the clarification. I really appreciate it. – user8081591 Aug 15 '18 at 22:38