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?
4 answers
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.
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?
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?
and would the answer be positive because the temperature is increased?
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.
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.
1 answer