[stardancer]

the molar heat of vaporization of benzene is 30.88KJ/mol. the boiling point of benzene at 1 atm is 78.3 degrees C. Calculate entropy change when 11.5 g of benzene boils at this temp

Delta S = delta H/ T
 = (30.8KJ/mol) (1000J/1 KJ)/ (78.3 + 273)K
 = 87.7 J/K mol
since this is for a mole of benzene we have to find how many moles are in 11.5 g
11.5g/(78.108g/mol) = 0.147mol
(87.7 J/k mol) * 0.147mol
 = 12.91
is this even being right? or am I off in lala-land?

[rawratthis]

your answer is reasonable J/K

[stardancer]

thank you very much!