- #1

jamalm55

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## Homework Statement

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?

## Homework Equations

chemical potential (mu) of any solution: μ=RTlna

_{i}

a

_{i}= γ

_{i}χ

_{i}

lna

_{A}= γ

_{A}χ

_{A}= -χ

_{B}

A= solvent, B= solute

## The Attempt at a Solution

So since its asking for Δμ, I put Δμ

_{sucrose}-Δμ

_{saccharin}which simplifies to Δμ=RT(lna

_{sucrose}-lna

_{saccharin}).

Then I found moles of each substance from the given info: n

_{H2O}=13.71 mol

n

_{sucrose}= 0.022 mol

n

_{saccharin}= 5.46 mol

Then I found their mole fractions, χ

_{B},

χ

_{B}=χ

_{sucrose}= n(sucrose)/(n(H2O)+n(sucrose))= 1.60 x 10^-3

χ

_{B}=χ

_{saccharin}= n(saccharin)/(n(H2O)+n(saccharin)) = 0.285

And plugged into Δμ=RT(lna

_{sucrose}-lna

_{saccharin}).

but I'm not sure if I have to sub in ln(χ

_{suc}) - ln(χ

_{saccharin}) or just

-χ

_{sucrose}- (-χ

_{saccharin})

I figured since lna

_{A}= γ

_{A}χ

_{A}= -χ

_{B}, 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.