The heat of vaporization of alcohol is 879 J/g. What is the energy in J required to vaporize 4.50 g of #C_2H_5OH#?

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Answer 1 · 2017-05-27T07:32:35

This simply the product #Delta=DeltaH_"vaporization"xx"amount of alcohol."# #~=4000*J#.

Practically, this process described the process:

#"Ethanol(l)"+Deltararr"Ethanol(g)"#

The ethanol would have to be at its normal boiling point.........

And so, we calculate..................

#4.50*gxx879*J*g^-1=??*J#