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chemistry - Molar enthalpy

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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?

  • chemistry - Molar enthalpy -

    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 -

    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?

  • chemistry - Molar enthalpy -

    and would the answer be positive because the temperature is increased?

  • chemistry - Molar enthalpy -

    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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