[pchem]

We start with n moles of supercooled h20(l) at 3 degrees celcius in a box insulated fromt he surroundings at a pressure of 1 bar.    Some of the water is allowed to freeze and the whole system is brought back to 0 degrees celcius (where liquid and ice coexist in equilibrium.  This process is conducted at constant pressure and adiabatic conditions.  

Q: at this poin what is the fraction n(liquid)/n(total) of moles in the liquid state.

Information: DHfusion=3000 J/mol.  Cm,p(ice)=28 J/mol*K,  
Cm,p(liquid)=50J/mol*K

What I tried to do is use the formula DS=integral(nCm,pdT)-Dfusion+integral(nCm,p dT/T(ice)

i pretty sure that's completely worng.  can someone lend a hand pleaes?

[Demotivator]

(I assume you mean -3 C for the supercooled water.)
This is adiabatic so the sum of all heat processes equals 0.
That means just enough ice forms that the heat that is evolved is absorbed to raise the temp of liquid and ice 3 degrees to 0 C, so that there is no net change in heat.
now,
n = total moles
n(ice) = n-n(liquid)
H = heat of fusion

0 = (heat evolved as H) + (heat absorbed using Cice) + (heat absorbed using  Cliquid)
0 = -n(ice)H + n(ice)Cice(deltaT) + n(liq)Cliq(deltaT)
0 = -(n-n(liq))H + (n-n(liq))Cice(3) + n(liq)Cliq(3)
0 = -H + Cice(3) + [n(liq)/(n-n(liq))]Cliq(3)
[n(liq)/(n-n(liq))]=  (H - Cice(3))/Cliq(3)
n(liq)/(n-n(liq)) = [3000 - 28(3)]/50(3) = 19.44

n(liq) = 19.44(n-n(liq)) = 19.44n - 19.44n(liq)
20.44n(liq) = 19.44n
n(liq)/n = 19.44/20.44 = 0.95