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Topic: Heat of fusion...  (Read 2956 times)

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Offline 80z_chick

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Heat of fusion...
« on: October 10, 2012, 09:25:40 PM »
Hi everyone.  I am an older student who is taking grade 12 chemistry through correspondence in order to apply to college.  Unfortunately, math was not my strong point, and there's a lot of math involved with chemistry!  Could I get some help with this question?

"If ΔHFUS for NaCl is 38.0 kJ/mol, how much energy is required to melt 350 g of NaCl at its melting point?"

So, here's what I have so far:
ΔH = 350g NaCl x 1mol/58.44 x 38 kJ/mol
ΔH = 350 g x 0.0171 x 38 kJ/mol
ΔH = 167.7 kJ
So.. the melting point of NaCl is 801 degrees c.  Where does this fit into the equation?  The 58.44 is the molar mass of NaCl (I got this number from the internet). Am I on the right track at all with this? Any help would be greatly appreciated.  Thanks!

Offline curiouscat

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Re: Heat of fusion...
« Reply #1 on: October 11, 2012, 02:05:32 AM »
To me it looks like you have the answer already. But I'll wait for someone better at this to confirm.

Offline Borek

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Re: Heat of fusion...
« Reply #2 on: October 11, 2012, 04:31:49 AM »
Melting point is unrelated to the question. Your approach is correct, but somehow you managed to give a wrong answer. Check your calculations.
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Offline fledarmus

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Re: Heat of fusion...
« Reply #3 on: October 11, 2012, 09:07:09 AM »
Hmm, when I plug those numbers into my calculator, I don't get the same answer you do...


Offline 80z_chick

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Re: Heat of fusion...
« Reply #4 on: October 11, 2012, 08:12:38 PM »
Oops.. my bad.  How about 227.4 does that sound better?  So why do they even mention the melting point in the question if it's irrelevant?

Offline fledarmus

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Re: Heat of fusion...
« Reply #5 on: October 11, 2012, 08:18:34 PM »
Because, by definition, the heat of fusion is the energy required to melt the substance at its normal melting point. If you need to calculate the heat required to melt the compound at any other temperature, you have to include the contributions of energy required to raise the substance to its melting point or lower it from its melting point as part of the calculation.

And yes, that was the answer I got  :D

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