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 February 17th, 2018, 11:36 PM #1 Newbie   Joined: Feb 2018 From: Iran Posts: 16 Thanks: 3 Heat of reaction KOH(s) → K+(aq) + OH– (aq) ΔHdissolution = –56 kJ The temperature of 150 mL of water rises by 26.76 °C after adding some amount of KOH. What is the molar concentration of the solution? (cwater= 4.186 J/g °C, dwater = 1 g/mL) (neglect change in the volume by the addition of KOH)
 February 18th, 2018, 01:14 AM #2 Senior Member   Joined: Jun 2015 From: England Posts: 853 Thanks: 258 Perhaps something was lost in translation? What is the molar concentration of what? My guess would be they are asking for the molar concentration of the potassium ion, but they may be basing the question on potassium hydroxide. Secondly your enthalpy of dissolution should be in kJ per mole.
 February 18th, 2018, 01:30 AM #3 Senior Member   Joined: Jun 2015 From: England Posts: 853 Thanks: 258 So let us proceed as follows, working on the mole fraction of potassium hydroxide. Molecular mass of KOH = 56 Molecular Mass of water = 18 So mass of 150 ml of water = 150 x 1 grams or 150/18 moles. Now consider the heat. Heat added to water = Mass water x specific heat x temp rise = (150 x 4.186 x 26.76)J 1 mole KOH provides 56000 J So 1/56000 mole KOH provides 1 J So (150 x 4.186 x 26.76)/56000 mole KOH provides (150 x 4.186 x 26.76)J This was the number of moles of KOH added. Molar concentration = mole fraction = moles KOH / total moles = moles KOH/ (moles water + moles KOH) Can you finish this now? Thanks from Elize

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