Which has the highest boiling point and lowest freezing point?

Question

(a) 0.4 m NaCl

0.3 m Na2SO4

0.5 m sucrose (nonelectrolyte)


(b)0.2 m HCl

0.2 m acetic acid

0.3 m KOH (weak electrolyte)

(c) 0.5 m ethylene glycol (nonelectrolyte)

0.4 m glucose (nonelectrolyte)

0.3 m BaCl2



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(a) Highest BP: .4m NaCl
Lowest FP: .5 m surcrose

(b) Highest BP: ..3m KOH
Lowest FP: .2m acetic acid

(c) Highest BP: .3m BaCl2
Lowest FP: I am unsure

Could someone check this for me?? Thank you!!!

Answer 1

How did you arrive at these answers. If you used
delta T = i*Kb*m and
delta T = i*Kf*m it seems to me that the material having the highest boiling point and the material having the lowest freezing point must be the same? And you have two answers for each. The secret is i*m.

Answer 2

By the way, KOH is not a weak electrolyte. Just the contrary.

Answer 3

a) the one that has the highest bp (and lowest fp, and remember, that solution will be both. Lowest FP means greatest fp depression). Doesn't the effective molality i*m depend on the number of particles, NaCl is 2, Na2So4 is three, and sucrose is one. I don't agree with your answer.
b) HCl completely (almost) dissociates, into 2 particles per molecule, the others hardly dissociate at all.
c)BaCl2 breaks into 3 particles per molecule

Answer 4

Im really sorry. Im actually looking for the highest and lowest boiling points. I typed it wrong.

Answer 5

ok. So i*m is still the secret.
For a.
NaCl = 0.4 m. i=2 i*m=0.8
Na2SO4 = 0.3m. i=3 i*m=0.9
sucrose = 0.5m. i=1 i*m=0.5
So the highest boiling point will be the one that has the highest i*m which is Na2SO4. The lowest boiling point will be the lowest i*m which is sucrose.
Part b and c are done the same way.
b is correct.
c is not.
i for ethylene glycol is 1.
i for glucose is 1.
i for BaCl2 is 3.