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Homework Help: Difference in chemical potential(mu) problem

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Given: 2 cups of hot coffee, each 250 mL, T=50 degrees celsius, 323.15K
    1 cup is sweetened 1tsp (5cm^3) of sucrose (342.3 g/mol)
    the other cup is sweetened with 1g saccharin (183.17 g/mol)
    assume coffee is very weak so the sweetener is the only solute
    density of H2O at 323.15K = 988.07 kb/m^3
    density of sucrose = 1505.5 kg/m^3
    what is the difference between the chemical potential of the hot water
    in the two cups?

    2. Relevant equations
    chemical potential (mu) of any solution: μ=RTlnai
    ai = γiχi
    lnaA = γAχA = -χB
    A= solvent, B= solute

    3. The attempt at a solution
    So since its asking for Δμ, I put Δμsucrose-Δμsaccharin which simplifies to Δμ=RT(lnasucrose-lnasaccharin).

    Then I found moles of each substance from the given info: nH2O=13.71 mol
    nsucrose= 0.022 mol
    nsaccharin= 5.46 mol

    Then I found their mole fractions, χB,
    χBsucrose= n(sucrose)/(n(H2O)+n(sucrose))= 1.60 x 10^-3
    χBsaccharin= n(saccharin)/(n(H2O)+n(saccharin)) = 0.285

    And plugged in to Δμ=RT(lnasucrose-lnasaccharin).
    but I'm not sure if I have to sub in ln(χsuc) - ln(χsaccharin) or just
    sucrose - (-χsaccharin)

    I figured since lnaA = γAχA = -χB, I do the latter.

    Δμ=(8.314 J/molxK)(323.15K)((-1.60x10^-3)-(-0.285)) = 761.4 J/mol

    I can't find the solution anywhere to check.
  2. jcsd
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