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, χB=χsucrose= n(sucrose)/(n(H2O)+n(sucrose))= 1.60 x 10^-3 χB=χsaccharin= 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 -χsucrose - (-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.