Answers, In-class or in-lab Exercises, Solutions
1. A sample of caffeine, C8H10N4O2, was dissolved in 45.0 g of chloroform, CHCl 3, to give a 0.0946 m solution.
How many grams of caffeine were in the sample? MM(caffeine) = 194.1 g/mol.
Answer: 0.826 g caffeine
Strategy:
0) I always write out m as “moles solute / kg of solvent”, just as I always write out M as “moles of solute /
L of solution” so that I can better “see” what the units “mean”. It is easier to move forward then using
either “rationalization” or “dimensional analysis” (or both)
1) Recognize that this problem describes a solution with a specific amount of solvent (and thus a specific
amount of solute)
2) Use that given amount (here, 45.0 g of CHCl3), along with the molality of the solution (moles caffeine
per kg CHCl3) to find the moles of caffeine in the solution. (You should recognize that g of CHCl3
needs to be converted to kg since the denominator of the molality is kg, not g of solvent).
3) Realize that moles of caffeine is just one step away from grams! Use the molar mass provided to
calculate grams from moles.
Execution:
45.0 g CHCl 3 x
1kg
0.0946molescaffeine 194.1g caffeine
x
x
 0.82628..  0.826g caffeine
1000 g
kg CHCl 3
mol caffeine
2. “Molality” is “moles of solute per kg of solvent” and “molarity” is “moles of solute per L of solution”. “Mass
percent” (of solute in a solvent) is “grams of solute per gram of solution” x 100. A 0.688 m citric acid solution
has a density of 1.049 g/mL. The molar mass of citric acid is 192.12 g/mol.
(a) Find the mass % of citric acid in the solution. Show your work.
Answer: 11.7%
Strategy:
0) Again, write out 0.688 m as 0.688 moles citric acid / kg H2O.
1) Write out the definition for the unit asked for:
mass % =
mass of solute
mass of citric acid
x 100 
x 100  you need to get the “numerator” and the
mass of solution
mass of solution
“denominator”
2) Recognize that this problem does NOT describe any specific amount of either solute, solvent, or
solution. Thus, you can choose a convenient amount of any one of these (and then move forward).
3) Recognize that it will be best for this problem to choose EITHER a mass of solute OR a mass of
solvent (since the molality relates these two—so once you pick one, you can calculate the other). I find
it easiest to choose the value of a “denominator” [and then you’ll also “have” the numerator as well].
That is, here choose a solution with exactly 1 kg of H2O in it. Then that means it will have 0.688 moles
of citric acid in it (since the molality is 0.688 moles citric acid per kg of H2O.
4) To get the numerator (mass of citric acid), use the molar mass to calculate grams from moles.
5) To get the denominator (mass of solution), realize that the mass of the solution is equal to the mass
of citric acid + the mass of the solvent). Both of these quantities are already known (once you’ve
done step 4 [because you chose 1 kg of H2O, right? Just remember to convert kg to g so the units are
the same].
6) Plug values into the definition and calculate.
Execution:
mass % 
mass of citric acid
x 100
mass of solution
Assume a solution with 1 kg H2O. Then it must also have 0.688 moles citric acid.
mass of citric acid = 0.688 mol x
mass of H2O = 1 kg x
 mass % 
192.12g
 132.18 g (citric acid)
mol
1000 g
 1000 g (H 2 O)
kgl
132.18
x 100  11.67...  11.7%
132.18  1000.g
(b) What is the molar concentration of citric acid in the solution? Show your work.
Answer: 0.638 M citric acid (aq)
Strategy:
0) Realize that you’ve already done work for this problem in part (a). If you had not done that, you’d do
steps similar to those describe in (a).
1) Write out the definition for the unit asked for:
molarity =
mol of solute mol of citricacid
 you need to get the “numerator” and the “denominator”

L solution
L solution
2) Using the same solution as you chose in (a), you already “have” the moles of citric acid (=0.688)
3) You need the volume of the solution (in L). Recognize that you have the mass of the solution from (a)
(132.18 + 1000. g = 1132 g)
4) Use the density of the solution, along with the mass, to calculate the volume (in mL). Convert to L.
5) Plug values into the definition and calculate.
Execution:
molarity 
mol of citric acid
L solution
Assume a solution with 1 kg H2O. Then it must also have 0.688 moles citric acid. (Part (a))
moles of citric acid = 0.688 mol
mass of solution = 1132 g
Volume of solution = 1132 g solution x
molarity 
1mL of solution
1L
x
 1.0791L
1.049 g solution 1000 mL
0.688 mol citric acid
 0.63755..  0.638 M citric acid(aq)
1.0791 L solution
These problems should be done in pairs or groups in lab. If problems 1 and 2 were not completed in class, do
those problems as well, making sure that all students in the group understand how to do them.
3. NOTE: In this problem a “particle” is defined as EITHER a molecule or an ion of any kind.
(a) Which solution has the greatest concentration of “particles” in it?
(i) 0.2 m FeCl3(aq)
(ii) 0.7 m CH3OH
(iii) 0.3 m MgCl2
(iv) 0.4 m KCl
Reasoning:
FeCl3, MgCl2, and KCl are soluble ionic compounds and are thus strong electrolytes. CH3OH is a
molecular non-acid (or non-base) so it is a nonelectrolyte. So
Each mole of FeCl3 in solution ionizes into one mole of Fe
3+
-
ions and three moles of Cl ions:
FeCl3(aq) → Fe3+(aq) + 3 Cl-(aq) [= four moles of particles]
Each mole of MgCl2 in solutions ionizes into three moles of ions:
2+
MgCl2(aq) → Mg
(aq)
-
+ 2 Cl (aq) [= three moles of particles]
Each mole of KCl in solutions ionizes into two moles of ions:
KCl(aq) → K+(aq) + Cl-(aq) [= two moles of particles]
Each mole of CH3OH does not ionize at all, yielding only one mole of molecules in solution:
CH3OH(aq) → CH3OH(aq) [= one mole of particles]
Thus:
0.2 m FeCl3(aq) contains 0.2 x 4 = 0.8 m particles
 0.2 molesFeCl 3 4 molesions 0.8 molesions

x

 0.8 m particles

kg
H
O
1
mole
FeCl
kg
H
O
2
3
2


0.7 m CH3OH(aq) contains 0.7 x 1 = 0.7 m particles
0.3 m MgCl2(aq) contains 0.3 x 3 = 0.9 m particles ← GREATEST CONCENTRATION OF PARTICLES
0.4 m KCl(aq) contains 0.4 x 2 = 0.8 m particles
(b) It turns out that the magnitude of the freezing point change or boiling point change induced by the
dissolution of a solute in a liquid is proportional to the total molality of particles in the solution (regardless
of what “kind” of particle or the size per particle). That being the case, which solution above has the
highest boiling point?
Answer: (iii) 0.3 m MgCl2
Reasoning: The boiling point of a liquid is elevated when solute is present. The solution with the highest
concentration of (total) particles will have its boiling point elevated the most. From (a), that solution is 0.3
m MgCl2.
(c) Which solution above has the highest freezing point? Be careful to read this question carefully!
Answer: (ii) 0.7 m CH3OH
Reasoning: The freezing point of a liquid is depressed (i.e., lowered) when solute is present. The
solution with the highest concentration of (total) particles will have its freezing point depressed the most,
and thus will have the lowest freezing point. This question asked which solution has the highest freezing
point, which means the one whose freezing point has depressed the least. That will be the one with the
smallest concentration of total particles. Thus it is the CH3OH solution (see values in (a)).
4. If 500. g of methanol (CH3OH) is added to 2.1 kg of water to form a solution, by how many degrees would
the freezing point be lowered? Kf for water is 1.86 C/m.
Answer: 14 °C
Work/Reasoning:
Since Kf is 1.86 C/m, we just need to know the molaltiy of the solution to calculate the number of
degrees that the freezing point lowered. Molality = moles solute per kg of solvent, so calculate the
moles of CH3OH from the grams using the molar mass (calculate it from the molecular formula). Then
put that over the 2.1 kg of water (the solvent) to calculate m. Then multiply by 1.86 C/m.
Molar mass of CH3OH = 12.01 + 3(1.01) + 16.00 + 1.01 = 32.05 g/mol
 500. g x
molality
Tf |=
1mol
 15.60 mol (reasonable: 500 g is a lot bigger than 32 g)
32.05 g
molessolute molesCH3 OH 15.6 molesCH3 OH


 7.428..m
kg solvent
kg H2 O
2.1kg H2 O
1.86 C
x 7.428 m  13.8  14 C
m