How many grams of methanol could be vaporized by the addition of 22.45kJ of heat?

Use the formula

q = m·ΔHv

where

q=heat energy=22.45kJ= 22.45 x 10^3J
m=mass=?
ΔHv=heat of vaporization=1,006 J/g

Solve for m,

q/ΔHv=m

(22.45 x 10^3J/1,006 J/g)=m