Question

At 1 atm pressure, the heat of sublimation of gallium is 277 kj/mol and the heat of vaporization is 271 kj/mol. to the correct number of significant figures, how much heat is required to melt 4.50 mol of gallium at 1 atm pressure?

Answer 1

Below are the choices that can be found elsewhere:

 a. 268 kJ 
b. 271 kJ 
c. 9 kJ 
d. 6 kJ

So the key thing to realize here is what the information given to you actually means. Sublimation is going from a sold to a gas. Vaporization is going from a liquid to a gas. Hence you can create two equations from the information that you have: 

Ga (s) --> Ga (g) delta H = 277 kJ/mol 

Ga (l) --> Ga (g) delta H = 271 kJ/mol 

From these two equations, you can then infer how to get the melting equation be simply finding the difference between the sublimation (two steps) and vaporization (one step). 

Ga (s) --> Ga (l) delta H = 6 kJ/mol 

At this point, all you need to do is a bit of stoichiometry. You start with 1.50 mol and multiply by the amount of energy per mole (6 kJ/mol). 

*ANSWER* 
9 kJ/mol (C)