Freezing Point Question?

Question

Which of the following aqueous solutions will have the highest freezing point temperature?

.100 molal MgBr2
.100 molal MgSO4
.150 molal KClO3
.150 molal MgCl2
.250 molal sucrose

If you could tell me why that would be helpful, too, thanks.

Answer 1

The freezing point depression is dependent on the amount of ions present that physically get in the way of the crystalization process.

MgBr2 dissociates into 3 ions, .3 moles of ions
MgSO4 dissociates into 2 ions .2 moles of ions
KClO4 dissociates into 2 ions .3 moles of ions
MgCl2 dissociates into 3 ions .45 moles of ions
sucrose does not dissociate .25 moles of ions

MgCl2 will depress the freezing point the most with .45 moles of ions

Answer 2

Freezing point depression

Delta Tf = Kf x m x i

Kf is the freezing point constant for the solvent - it will be the same value for ALL the species given, so if we didn't know the solvent we could make up a number. Since these are aqueous species that means water so a Kf value of 1.86 oC/m

m = concentration (molality - which those are given to you)

i is the van't Hoff factor. Ideally, when these species are put into a solvent, how many particles are formed? (Covalent species stay as ONE particle i = 1, while ionic species break up into ions

0.100 molal MgBr2 x 3 ions x 1.86oC/m = 0.558 oC so the new f.p. of this solution would be the normal f.p. of water (0 oC) - the delta T that we calculate
-0.558oC

MgSO4: 0.100 m x 2 ions x 1.86 oC/m = 0.372 oC

new f.p of water = -0.372 oC

KClO3: 0.150 m x 2 ions x 1.86oC/m = 0.558 oC

new f.p. of water = -0.558 oC

MgCl2: 0.150 m x 3 ions x 1.86oC/m = -0.837 oC

new f.p of water = -0.837 oC

sucrose: 0.250 m x 1 particle x 1.86 oC/m = 0.465 oC

new f.p. of water = -0.465 oC

Thus, the MgCl2 lowers the f.p. of water the most. Pervious posters are correct, the MgSO4 guy is wrong!!

Answer 3

the highest freezing point is for the compound who has the LOWEST depression
This is for the compound which has the lowest osmolarity

.100 molal MgBr2 osmolarity =.100*3 = 0.300
.100 molal MgSO4 osmolarity =.100*2=0.200
. 150molal KClO3 osmolarity =.0.150*2 =0.30
.150 molal MgCl2 osmolarity =.0.15*3=0.45
.250 molal sucrose osmolarity=0.25

so the answer is .100 molal MgSO4 which has a depression of 0.2*0.186 =0.372 and freezees at -0.372C

Answer 4

.150 molal MgCl2 will have highest freezing point because the vant hoff factor(i) is 3.
Depression in freezing point=i*molality*Ebuli s constant
so by your statistics sucrose will have highest freezing point.

Answer 5

MgCl2

Because i=3, and freexing point depression is a colligative prperty, therefore, substance is not as important as the quantiity.