From data below, calculate the total heat (J) needed to convert 0.172 mol of gaseous ethanol (C2H6O) at 351° C and 1 atm to liquid ethanol at 25.0° C and 1 atm.

Question

Boiling point at 1 atm 78.5°C
c gas 1.43 J/g · °C
c liquid 2.45 J/g · °C
ΔH°vap 40.5 kJ/mol

My work:
.172 mol = .003734 g
1) q = mc(T2-T1)= (.003734)(1.43)(351-78.5) = 1.455 J
2) q = n*delta Hvap =.172*40500 = 6966J
3) q = mc(T2-T1) = (.003734)(2.45)(78.5-25) = .4894 J
-------------- adding all the q together gives me -6.97e3 J, its negative since im going from g to l. Its also 3 sig figs. The computer tell me my answer is off by 10%. So what am I doing wrong. Someone please tell me or I will become insane and fail my chem test then drop out of college which will lead me to become homeless, all because of this stupid problem!

Answer 1

Is 0.172 mol ethanol really 0.003734 g? Isn't it 0.172mol x 46 g/mol = 7.912 g?

If you still have a problem, post the boiling point and freezing points of ethanol. I wouldn't want you to become homeless. ;-)

Answer 2

Your plight reminds me of this old proverb:

"For Want of a Nail

For want of a nail the shoe was lost.
For want of a shoe the horse was lost.
For want of a horse the rider was lost.
For want of a rider the message was lost.
For want of a message the battle was lost.
For want of a battle the kingdom was lost.
And all for the want of a horseshoe nail."

Answer 3

And for want of an answer the home was lost.