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posted by  Vanity Duluoz on 9/11/2008 8:42:43 PM  |  status: Live  

Chemistry II: solutions, change temp, heat of solution

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Question Details:
40.12 grams of a substance with a molar mass of 493.714 g/mol are dissolved in 338.21mL of water. The temperature of the system changes from 70.07 to 36.36 degrees C during the mixing. What is the heat of solution (ΔH) for this process in kJ/mol? 
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posted by Ricky Wilkins on 9/12/2008 5:16:24 AM  |  status: Live
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Vanity Duluoz's comment:
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Response Details:
 We know that the heat change for areaction
                                     q = msΔt
 Where m -mass of the substance = 40.12g
            s - specific heat capacity of the water = 4.184J/g 0C
           Δt = tfinal - tinitial              = 36.360C - 70.070C
                                                   = -33.710C
 ∴ q  = 40.12g * 4.184J/g 0C * -33.710C
         = -5.65 kJ
 Therefore ΔHsoln     =  -q / mol solute
 number of mols of solute = 40.12g / 493.714g/mol
                                       = 0.08126 mol
                        ΔHsoln     =  -(-5.65 kJ) / 0.08126 mol
                                       =  69.529 kJ/mol
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posted by Ainala8055 on 9/12/2008 6:53:22 AM  |  status: Live
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Vanity Duluoz's comment:
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Response Details:
                 q = msΔt
            Δt = tfinal - tinitial  = -33.710C
 ∴ q  = 40.12g x 4.184J/g 0C x -33.710C
         = -5.65 kJ
 mols of solute = 40.12g / 493.714g/mol
                                       = 0.08126 mol
                        ΔHsoln     =  -(-5.65 kJ) / 0.08126 mol
                                       =  69.529 kJ/mol
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