# The heat of vaporization of alcohol is 879 Jg^-1. What is the energy (in J) required to vaporize 4.50 g of alcohol, C_2H_5OH?

$879$ joules per gram times $4.50$ grams equals $3955.5$ joules
This is a fairly straightforward question. Each gram of alcohol requires $879$ $J$ to vaporise it, and there are $4.5$ $g$, so we simply multiply.
Notice the units: $J {g}^{-} 1 \cdot g = J$