What happens to a solution below its freezing point?

This is the question we did in class-


1000g of 1 molality sucrose solution in water is cooled to -3.534°c. What mass of ice would be seperated out at this temperature? (Kf of H2O= 1.86 K.mol^-1.kg)


We first found the lowering of freezing point (FP) at 1 molality concentration of sucrose which will be 1.86°c. Then from the data of mass of solution to be 1000g we found the mass of solute as 254.84g and mass of solvent as 745.15g. Then we applied the formula of delta(Tf)= Kf\* molality, again. (Note- delta(Tf) is the lowering in FP). This time the moles of solute would remain the same so we put that as 254.48/342. The new mass of solvent was taken 'x'(g) so the molality was [(254.84)(1000)]/[(342)(x)]. Delta(Tf) was taken as 3.534°c. Solving the equation we found the value of 'x' and the difference of 745.15-x was taken as the mass of ice seperated out.


Now, I understand the solution till we find mass of solute and solvent. But after that why would we take lowering in FP as 3.543? Why would less ice be obtained if the temperature is even below the FP of solution? I mean shouldn't all the solvent be solidified? Why plug in that formula again?
Can someone please make this clear or give an alternate solution.




Accept some experimental facts and then proceed to calculations.


a)First of all, as you know very well that 1 m (molal) sucrose solution will not freeze at 0 C.


b) Additionally, you have also calculated that 1 m sucrose solution has 254.84 g sucrose and 745.15g water. This is the solution before cooling.


c) Now the question mentions that that the solution has been cooled to -3.534 C. Again, experimentally, it may be observed that some *pure* water freezes out and some liquid containing sucrose remains there.


When (c) happens, there will be a *change* in molality of sucrose. Now you have a biphasic system, solid pure ice and sucrose solution, you now would start your calculations.


Under the experimental conditions $\Delta$T= 3.534, this *temperature change* corresponds to a sucrose molality of = 3.534= K$\_f$.m.


Therefore the molality of the remaining liquid sucrose solution at -3.5 C must be 1.88 m! The sucrose solution has become concentrated now pure water has frozen out.


From 1.88 m you calculate the amount of water in the liquid. You also know the original amount of water you started with before cooling. This should help you calculate the amount of ice.


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