Asked by Rose Bud

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?

i tryed

q=(m)x(c)x(delta T)

didn't really work. i got 22.35 J

do i use that to find the molar enthalpy if so how?
Answered by DrBob222
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.
Answered by Rose Bud
haha, yep,

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?
Answered by Rose Bud
and would the answer be positive because the temperature is increased?
Answered by DrBob222
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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