[oceanmd]

Please solve this problem. Thank you very very much
22.0 grams of liquid benzene at 8.0 degrees C is heated to gaseous benzene at 105.0 degrees C. What is the total amount of heat absorbed in this process? (Hvap =567J/g, Cliq=1.69 J/gC, Cgas = 1.05 J/gC

My solution:
567 J/g of energy is absorbed when 1 mole of benzene is converted to one mole of benzene vapor.
22.0 g of benzene = 22.0 /78.114 = 0.282 mol C6H6 (I converted mass to mol)
q (heat) = 567 x 0.282 / 1 mol = 159.89 J/g

I have read the chapter on Changes of States, but there are no example problems to show how to do this kind of problem.
Please tell me what I am doing wrong. Should I be using the formula q=c x m x change of T?

Thank you
 
 

[enahs]

567 J/g of energy is absorbed when 1 mole of benzene is converted to one mole of benzene vapor.

No. 567 J of energy is absorbed for every 1 gram of benzene when being converted to vapor. Mols are not required for this question at all.

However, the boiling point is required.

Step A)
How much energy is required to raise 22 grams of benzene from 8^oC to whatever it's boiling point is?
Step B)
How much energy is required to vaporize 22 grams of benzene?
Step C)
How much energy is require to raise 22 grams of gaseous benzene (from whatever temperature that may be) to 105^oC?

[oceanmd]

Please check what I came up with:
Thank you very much.
Step A) Boiling Point is 80.1 degrees C
q = c x m x delta T
q = 1.69 x 22 x (80.1 - 8 ) = 2680 J

Step B) q = 22 x 567 = 12474 J
Step C) q = 1.05 x 22.0 x (105-80.1) = 575 J
Total: 2680 + 12474 + 575 = 15729 J = 1.573 x 10^4 J

[enahs]

Looks reasonable. I assume you plugged them into a calculator correctly. Correct method.

I would express it as kJ my self though.

[oceanmd]

Should I use kJ in all 3 steps instead of J? Thank you so very much for your help. I looked through the whole text book, it is not explained anywhere! Thanks!!!