chapter 12 solutions and colligative properties Flashcards
(11 cards)
What is the molality of a solution prepared by dissolving 0.1 moles NaCl in 11 moles of H2O?
Refer to the study guide. It’s not like I have this formula memorized or plan on memorizing it. But it’s:
molal = moles of solute / kg of solvent
a. 0.009m
b. 0.25m
c. 0.51m
d. 0.75m
e. 1.0m
c. 0.51 m
Just follow the formula.
You have to convert moles of H2O to grams and then convert grams to kg
What is the mass percent CH3OH for a solution prepared by adding 65g CH3OH to 450mL H2O? (density of water = 1g/mL)
This starts getting ridiculous because a little m in the question above stands for molal but a little m in this formula stands for mass. But anyway. The formula is:
mass % = mass of solute / mass of solution
a. 3.1%
b. 6.2%
c. 8.4%
d. 12.6%
e. 18.4%
d. 12.6%
This one tripped me up for a second. When it says ‘mass of solution’ in the denominator, it means the mass of the solution after you drop the CH3OH in it.
I guess otherwise, it would be the mass of ‘solvent’
What is the mass percent CH3CH2OH in 0.1m CH3CH2OH aqueous solution?
a. 0.46%
b. 5.0%
c. 11.4%
d. 27.2%
e. 61.4%
a. 0.46%
So this is really stupid but, in the study guide, little m stands for molal AND little m (in the mass% formula) stands for mass… Which really fucked me up. But anyway.
If the solution is 0.1 molal, that means that if your moles of solute and 1 then your kg of solvent has to be 10 since 1 / 10 = 0.1
From there, convert your moles on top (1 mole) to grams. Then add those grams to the denominator once you’ve converted the denominator’s kg’s to grams. You’ll get 45 divided by 10,045
Which of the following solutions would have the highest boiling point?
(little m does not equal mass in the boiling point formula, it equals molal)
a. 0.5m KCl
b. 0.5m CH3OH
c. 0.5m RbNO2
d. 0.5m MgCl2
e. 0.5m CsNO3
d.
The book’s answer makes sense. It’s just been a while since I’ve done this. Basically, for every Mg atom, there’s 2 Cl atoms so… 3
The greater the concentration of dissolved particles the greater the boiling point elevation. All these solutions have a 0.5m concentration; the difference is in the number of ions they dissociate into (the van’t Hoff factor, i).
KCl i=2
CH3OH i=1
RbNO2 i=2
MgCl2 i=3
CsNO3 i=2
As MgCl2 dissociates to form the most ions it will have the highest boiling point.
Which of the following solutions would have the highest boiling point?
a. 0.1m Zn(NO3)2
b. 0.2m NaI
c. 0.2m (NH4)2SO4
d. 0.25m CH3CH2OH
e. 0.1m HI
c.
I got it wrong because I saw the c. is all molecular type bonds but that, in itself, was kind of wrong. SO4 is a polyatomic ion so it doesn’t really follow the rules of molecular bonds having low boiling points. I did at least know that both a. and c. separate into 3 ions each.
The greater the concentration of dissolved particles the greater the boiling point elevation. The product of the molality and the van’t Hoff factor gives the total concentration of dissolved particles.
0.1m Zn(NO3)2 im = (3)(0.1m) = 0.3
0.2m NaI im = (2)(0.2m) = 0.4
0.2m (NH4)2SO4 im = (3)(0.2m) = 0.6
0.25m CH3CH2OH im = (1)(0.25m) = 0.25
0.1m HI im = (2)(0.1m) = 0.2
As 0.2m (NH4)2SO4 dissociates to form the most ions it will have the highest boiling point.
Which of the following solutions would have the lowest freezing point?
a. 1m CsBr
b. 1m C3H8O
c. 1m FeCl3
d. 1m NiSO4
e. 1m CuCl2
c.
Yay! Finally got one. I figured that high boiling points and low freezing points (meaning: you have to drop the temp down super far to freeze it) are kinda 2 sides to the same coin. When you think about it, the higher the concentration of NaCl (salt) in water, it’s obviously going to take some serious cold to freeze it since we all know that water melts ice. But anyway. FeCl3 splits into the most ions and since they’re all the same concentration…
The greater the concentration of dissolved particles the greater the freezing point depression. All these solutions have a 1m concentration; the difference is in the number of ions they dissociate into (the van’t Hoff factor, i).
CsBr i=2
C3H8O i=1
FeCl3 i=4
NiSO4 i=2
CuCl2 i=3
As FeCl3 dissociates to form the most ions it will have the lowest overall freezing point.
Which of the following solutions would have the highest freezing point?
a. 1m HBr
b. 1m HF
c. 1m Zn(NO3)2
d. 1m Fe(NO3)3
e. 1m (NH4)3PO4
b.
Took a guess. Got it right. I knew that a. and b. both break up into 2 atoms apiece so they’re pretty weak as far as freezing points go i.e. ‘higher’
But the reason why I guess HF instead of HBr wasn’t entirely sound. But here’s a good explanation.
The greater the concentration of dissolved particles the greater the freezing point depression. The solution with the highest freezing point will therefore be the one with the smallest depression (the one that is decreased the least). All these solutions have a 1m concentration; the difference is in the number of ions they dissociate into (the van’t Hoff factor, i).
1m HBr i=2
1m HF i<2
1m Zn(NO3)2 i=3
1m Fe(NO3)3 i=4
1m (NH4)3PO4 i=4
HF is a weak acid and only dissociates to a small extent and therefore will have the lowest van’t Hoff factor, the lowest concentration of dissolved species, the smallest freezing point depression, and therefore the highest overall freezing point.
What is the freezing point of a solution prepared by adding 5.85g NaCl to 100g H2O? (Kf,H2O = 1.86˚C/m)
a. 3.72˚C
b. 1.86˚C
c. -1.86˚C
d. -3.72˚C
e. -7.44˚C
d. -3.72 degrees C
I got it wrong but I don’t feel too terrible about it since I did figure some stuff out. The given equation is 1.86 degrees per molal. So… You first have to find the molal before you can do anything. Convert 5.85 grams of NaCl to moles (0.1 moles). Then convert 100 grams of H2O to kg (0.1 kg). Which gives you 1 molal. The part I didn’t realize was you still have to factor in the charge magnitude (i) which is 2 for NaCl
So it’s really 1.86 in the negative direction times 2 because of how many ions it’s going to break into. So 1.86 times 2 is 3.72 (again, in the negative direction) so -3.72
The normal freezing point of benzene (C6H6) is 5.5˚C. What would be the freezing point of a solution prepared by adding 21g C6H12 to 100g of benzene?
(Kf,benzene = 5.12˚C/m)
a. 1.86˚C
b. 0˚C
c. -0.465˚C
d. -1.86˚C
e. -7.3˚C
e. -7.3 degrees C
This one took me forever to figure out. You go along and figure out your molal but the weird part comes when you’re trying to figure out the i factor. I kept thinking it was 3 since a reduced C6H12 would look like CH6 which might have a magnitude charge of 3. But that’s not the case since it’s a molecular bond.
Book: The van’t factor (i) for C6H12 is 1 as it is a molecular compound and doesn’t dissociate into ions.
What is the boiling point of 1.5m HOCH2CH2OH in water? (Kb,H2O = 0.51˚C/m)
a. 100.25˚C
b. 100.51˚C
c. 100.77˚C
d. 101.25˚C
e. 101.53˚C
.
