Posted by Rose Bud on Wednesday, February 29, 2012 at 9:58pm.
When 10.0 g of KOH is added to 100.0 g water, the solution's temperature increases from (25.18 C) to (47.53 C). What is the molar enthalpy of solution for KOH?
didn't really work. i got 22.35 J
do i use that to find the molar enthalpy if so how?
- chemistry - Molar enthalpy - DrBob222, Wednesday, February 29, 2012 at 10:06pm
No, that's not what you obtained. I know it isn't. 22.35 = Tfinal - Tinitial = 47.53-25.18 = 22.35 and that doesn't come close to m x c x delta T.
- chemistry - Molar enthalpy - Rose Bud, Wednesday, February 29, 2012 at 10:17pm
ok so q= 10,276.53 J
and to get the answer in molar enthalpy should i divide q by the mole of KOH
10.0g KOH (1 mol KOH/ 56.105g) = 0.1782 mol KOH
10,276.53 J / 0.1782 mol KOH = 57,668.5J/mol
= 57.6 kJ/mol
Is this right?
- chemistry - Molar enthalpy - Rose Bud, Wednesday, February 29, 2012 at 10:18pm
and would the answer be positive because the temperature is increased?
- chemistry - Molar enthalpy - DrBob222, Wednesday, February 29, 2012 at 10:42pm
Close but no cigar.
You had better redo the q part. I think it's closer to 9300 J or so. The rest of it is as you have it. q/mols KOH = J/mol but it usually is reported in kJ/mol.
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