SOLUTIONS Class #1
Objective: Describing what solutions are,
how they form, and how are they’re
strength is measured.
Put away your cell phone, all you need is your Solutions
packet, pen or pencil, reference tables, and an open mind.
Get ready for lots of questions for everyone!
1. A solution is a homogeneous mixture.
2. A solution forms when a solute is dissolved into a solvent.
3. If you put sugar into water, the sugar is the ____________
- while the water is the __________________
4. When a solution holds the maximum amount of stuff it is a
___________________________ solution.
5. If there is less than the maximum amount of solute in
the solution it is said to be __________________________
6. Only a few substances can form solutions when hot (like sugar, or sodium acetate)
– then you can cool them down and they can still hold all that solute when colder.
These solutions are said to be
____________________________________
7. Most solutions you think about will be aqueous
(which means dissolved into _________________________)
8. But they can also be gases (ex: __________ is a solution)
or even solids (ex: ________________is a solution)
When you try to dissolve stuff into solution, there are 3 factors that will affect this rate (either
making it faster or slower.
DEMONSTRATE
Let’s watch 2 forms of the same salt (CaCl2) dissolve into room temperature water
Let’s watch Alka-seltzer dissolve into hot or cold water.
Let’s watch chocolate powder mix into some milk with and without stirring.
TALK: What was the affect of particle size? How about temperature? How about agitation?
WRITE: When you try to dissolve a solid into a solution, these are the 3 factors that
would affect the rate of solvation:
9. _______________________________________________________
10. _______________________________________________________
11. ________________________________________________________
What 3 factors affect the rate of a solute
dissolving into a solvent?
Higher temperature of solvent
Faster solvent, more dissolving action, faster juggling
Agitation or stirring
Increases kinetic energy, faster dissolving action
Excess surface area of solute
more surface on solvent allows more solvation, less “waiting” for a
chance to swim in the solvent
How much solute will dissolve into a solution?
It depends first on
 14. Actual solubility (table F)
 15. Temp. of solvent (usually hotter = more)
 16. Pressure (affects only gases)
usually hotter temps = higher solubility
17. How does Coca-Cola get carbon dioxide into water to make soda?
Is CO2 a polar or nonpolar molecule? Is H2O a polar or nonpolar molecule?
Start explaining...
Draw the CO2 molecule.
Then draw a water molecule.
Measuring the strength of a solution.
MOLARITY
(M)
18. Molarity is the expression of concentration of a
solution as measured by the number of moles of
solute in a liter of solution.
19.
M=
# moles solute
Liters of solution
20.
What is the concentration of a
one liter salt water solution
containing 58 g NaCl?
(don’t even start without the formula!)
20.
What is the concentration of a one liter
salt water solution containing 58 g NaCl?
(don’t even start without the formula!)
M=
# moles NaCl
Liters of solution
=
1.0 mole NaCl
1 Liter solution
21. This is a 1.0 Molar NaCl solution,
or this solution has a 1.0 M NaCl,
or the solution contains 1 mole NaCl/Liter
22. If you add 29.0 g NaCl to enough water to form
a 600.0 mL solution, what is it’s concentration?
(write a formula or else)
22.
If you add 29.0 g NaCl to enough water to form a
600.0 mL solution, what is it’s concentration?
(write a formula or else)
M=
# moles NaCl
Liters of solution
=
0.500 moles NaCl
0.6000 Liters solution
23. This solution is 0.833 M
This solution has a molarity of 0.833
It’s am 0.833 M NaCl aqueous solution
23. You put 74.0 g KCl solid into a flask. You fill the flask to 1600.0 mL,
what is the molarity of this solution?
Start with the formula, or you know what might happen!
Your answer can be written as: _________________________________________________
or
________________________________________ or
_____________________________________________
23. You put 74.0 g KCl solid into a flask. You fill the flask to 1600.0 mL,
what is the molarity of this solution?
74.0 g KCl = 2.00 moles KCl
1600.0 mL = 1.6000 Liters
Molarity is moles/Liter so, 2.00 moles/1.6000 L = 1.25 M
Your answer can be written as:
the Molarity of this solution is 1.25
or
it’s a 1.25 M KCl solution, or
this solution contains 1.25 moles KCl/L
24. Calculate the molarity of a 750 mL KCl(AQ) solution containing 148 g KCl.
Write a formula or else go directly to jail, do not pass GO!, do not collect $200
24.
Calculate the molarity of an aqueous
solution containing 148 grams KCl of
750. mL
M=
# moles KCl
Liters of solution
=
2.00 Moles KCl
0.750 Liters solution
This solution has a Molarity of 2.67, or
It’s a 2.67 M KCl(AQ)
How many grams of sodium nitrate are in a one
liter aqueous solution that is saturated at 10°C?
25.
25.How many grams of sodium nitrate are in a one
liter aqueous solution that is saturated at 10°C
TABLE G…
10°C NaNO3
water
100 X = 80,000
80 g
100 mL
Xg
1000 mL
X = 800 grams NaNO3
How do we figure out the
MOLARITY OF THIS SOLUTION?
How do we figure out the
MOLARITY OF THIS SOLUTION?
How many moles is 800. g NaNO3 ? Calculate the molarity of this solution.
800. g NaNO3
1
x
1 mole NaNO3
85 g NaNO3
= 9.41 moles NaNO3
9.41 moles
1.0 liters
= 9.41 M
Solutions Vocabulary to Memorize by Tomorrow
26. Solute – the stuff dissolved into the solvent of a solution (the salt of salty water)
27. Solvent – the part of the solution that the solute is dissolved into, the water of salty water.
28. Saturated – when a solution is holding the maximum solute at that temperature.
29. Unsaturated – when a solution is holding LESS THAN the maximum solute for that temp.
30. Supersaturated – a strange circumstance where a solution is able to hold more solute than
normally possible at a particular temperature, usually formed by cooling a
saturated solution slowly.
31. Table G – the solubility guidelines for 10 compounds over all liquid water temps.
32. Molarity – the measured concentration of a solution in moles/Liter units.
33. Molarity Formula: M = Moles of solute
Liters of solution
34. What units go into this formula (only) – MOLES of solute and LITERS not mL! of solution
Solutions Class #2
Objective: More practice with solution concentration,
different molarity problems; how to make a
solution properly (and wrong!),
and then –
how to make a more dilute the solution
with it.
35. (write a formula!)
Calculate the molarity of an aqueous solution
containing 259 g KCl in a solution with total
volume of 750. mL
35. Calculate the molarity of an aqueous
solution containing 259 grams KCl in a
solution with total volume of 750. mL
M=
# moles KCl
Liters of solution
=
3.50 Moles KCl
0.750 Liters solution
This solution has a Molarity of 4.67, or
It is a 4.67 M KCl(AQ)
36.
How many grams of sodium chloride are in a one
liter aqueous solution that is saturated at 90°C?
36. How many grams of sodium chloride are in a one
liter aqueous solution that is saturated at 90°C?
90°C NaCl
water
100 x = 40,000
40 g
100 mL
xg
1000 mL
x = 400 grams NaCl
37. What’s the molarity of this saturated solution
of NaCl(AQ) ? (First convert these 400 grams of NaCl into moles)
37.
What is the molarity of this saturated solution of NaCl(AQ) ?
(First convert these 400 grams of NaCl into moles)
400 g NaCl
1
X
1 mole NaCl
58 g NaCl
= 6.90 moles NaCl
6.90 moles = 6.90 M NaCl
(AQ)
M=
1 liter
Does it matter how large or small
your saturated sodium chloride solution is when it
comes to the molarity? Start explaining now.
38. Hard thinking now:
38. Hard thinking now: does it matter how large or small your saturated
sodium chloride solution is when it comes to the molarity?
Start explaining now.
No, the molarity is the mathematical
expression of MOLES PER LITER.
Any change in solution size will either
increase the moles and liters proportionally,
or
Decrease the moles and liters proportionally.
You ALWAYS have the same concentration.
39. How many grams of NaCl are required to form a
2.50 L of 0.900 M NaCl(AQ)? (a formula will guide you)
39. How many grams of NaCl are required
to form a 2.50 L of 0.900 M NaCl(AQ)?
M=
# moles NaCl
Liters of solution
0.900M =
1
# moles = 2.25 moles
# moles NaCl
2.50 Liters
NaCl is 58 g/mole, so,
Solve for # moles, hint:
put Molarity over 1,
then cross multiply
2.25 x 58 = 130.5 g = 131 g NaCl (3 SF)
40. Calculate the grams of KOH required to make a
3.20 Liter solution of KOH(AQ) with a
1.20 M concentration. (always write a formula)
40. Calculate the number of grams of KOH required to
make a 3.20 Liter solution of KOH(AQ) with a
1.20 M concentration.
M=
# moles KOH
Liters of solution
1.20 M = # moles KOH
1
3.20 Liters
3.84 moles KOH
1
x
# moles KOH = 3.84 moles KOH
56 g KOH
1 mole KOH
= 215 g KOH
41. Quick reminder, a solution by definition is a ____________ ___________
Describe the RIGHT WAY to mix up a one liter 1.00 M NaCl aqueous solution?
42. The right way is to put _______________ into a beaker,
then add the ______________________________________
43. The WRONG WAY to make this solution would be to start with
____________________________ of ____________, and them add in the
________________.
If we do this WRONG WAY, our solution will have a
slightly higher _________________________, and it would be wrong.
44. How would you mix up a 2.50 M KNO3(AQ)
of 5.65 liters?
44. How would you mix up a 2.50 M KNO3(AQ) of 5.65 liters?
You would have to calculate the number of
moles of KNO3 you need:
2.50 moles X 101 g/mole = 253 grams (3 sf)
Put that into a beaker,
then fill up to the 5.65 Liter mark.
45. Using this 2.50M KNO3(AQ) how would you mix a KNO3(AQ)
that’s only 1.15 M and 1.64 liters in volume?
To do this we need a new formula, that’s not on our reference
tables. Let’s fix the reference tables, and come back to this in
a few minutes.
We will have to use some stock and some water to mix up the perfect new
solution. There is a formula called
the ____________________ ________________________ which is MISSING on
the reference tables.
Take out table T now, let’s FIX THIS immediately. Add the blue…
For mixing solutions from
scratch, or measuring
strength of solutions.
Concentration
Moles of solute
Molarity = Liters of solution
For mixing up a new solution
from a stock solution you have
on-hand
M1V1 = M2V2
Now let’s figure out what this formula
means, and how it will help us with this
problem.
This requires the dilution formula + what do all those letters & numbers mean.
45. The formula is: ___________________________ = _____________________________
The symbols mean:
46. ____________ is ________________________ ___________________________
47. ____________ is _____________________________ ______________________
48. ____________ is ______________________________ _________________
49. ____________ is ____________________________ ___________________
Problem I.
Using this 2.50M KNO3(AQ) how would you mix a KNO3(AQ)
that’s only 1.15 M and 1.64 liters in volume?
This requires the dilution formula and what all those letters + numbers mean.
The formula is:
M1V1 = M2V2
The symbols mean:
M1 is Molarity of the original stock solution
V1 is Volume of the original stock solution (the most common unknown)
M2 is Molarity of the new solution you want to make
V2 is Volume of new solution that you want to make
50. The solution that you START with is called the
_______________________
SOLUTION
quite literally because that is what you have
______
!
__________________
51. Your NEW solution must always be _________ __________________
than the starting solution
52. Using this 2.50M KNO3(AQ) how would you mix a
KNO3(AQ) that’s 1.15 M and 1.64 liters in volume?
(finally)
52. Using this 2.50 M KNO3(AQ) stock solution, how would you mix a KNO3(AQ)
that’s 1.15 M and 1.64 liters in volume?
M1V1 = M2V2
(2.50 M)(V1) = (1.15 M)(1.64 L)
V1 =
(1.15 M)(1.64 L)
2.50 M
V1 =0.754 Liters of stock
You need 0.754 L (754 mL) of stock, but
that’s NOT how to make this solution.
54. FIRST: put that 754 mL
stock into a beaker
SECOND: fill up with pure
water to the 1.64 L mark
(1640 mL) because that is the
TOTAL VOLUME of solution
you want to make.
Your solution is part stock, and
part water.
55. How do you prepare a 135 mL NaCl(AQ) solution
of 1.00 molarity from a stock solution of 5.50 M?
M1V1 = M2V2
55. How do you prepare a 135 mL NaCl(AQ) solution
of 1.00 molarity from a stock solution of 5.50 M?
M1V1 = M2V2
(5.50 M)( X mL) = (1.00 M)(135 mL)
(1.00
M)(135
mL)
x mL =
5.50 M
x = 24.5 mL stock solution
(3 SF)
But is that (?) the answer???
Short answer: NO. It’s not enough. Let’s think.
We want a 135 mL solution, this math tells us
HOW MUCH STOCK solution to start with. So…
2. Fill up to exactly
135 mL with water.
(135 – 24.5 = 110.5 mL water)
water
Stock solution
1. Put your 24.5 mL STOCK
SOLUTION into a new beaker.
56. Hydrochloric acid solution comes to the school very concentrated –
at 12.0 M! We can’t use that in lab, it’s too dangerous and too strong
for any high school chem lab work. Using the stock solution, exactly
what should your teacher do to make up 2.00 L of 2.25 M HCl solution
(which is also pretty strong!) start with a formula, it helps!
56. We buy 12.0 M HCl(AQ). How would you mix up
a 2.00 liter hydrochloric acid solution of 2.25 M?
M1V1 = M2V2
(12.0 M)(x L) = (2.25 M)(2.00 L)
x L = (2.25 M)(2.00 L)
12.0 M
X = 0.375 L stock solution
which = 375 mL
To mix this solution up, you would first take
exactly 0.375 Liters of stock HCl(AQ)
(that is 375 mL), and put it into a big beaker,
then you’d pour in enough water to dilute it to
2.00 liters (that is 2000 mL).
375 mL stock + 1625 mL water
= 2000 mL new solution
57. Once you have a solution you can make it less concentrated.
You can ______________________ it.
58. You can do this by __________________ more ________________.
In high school, and without doing a LOT
of math, you can’t make your stock
solution more concentrated and know
the concentration of the new solution.
You could just put more solute in, but you
could never figure out the math.
Solutions Class #3
Objective: Another way to measure low concentrations of
very weak solutions called PARTS PER MILLION,
+ working with the 3 colligative properties of water
You have a very weak salt water solution that contains just 1502 grams NaCl
dissolved into 312,000 liters of water. What is the molarity of this solution?
(the answer is stupidly small to make a point)
1502 g NaCl
1
X 1 mole NaCl = 25.9 moles salt
58 g NaCl
25.9 moles
312,000 Liters
M = 0.0000830 Molar NaCl(AQ)
That’s 840 ten millionths Molar!!
Sort of a stupid number.
It’s true, but the number is way too small to “make any sense” to us.
We need a different way to measure strength of really weak solutions,
called PPM
59. PPM stands for:
___________________________________________________________
The formula is on the back of your reference table, look for it now, copy
it into your notes here
60………………………
PPM =
61. You have a very weak salt water solution that
contains just 1502 g NaCl dissolved into 312,000 liters
of water. What is the molarity of this solution?
What is the concentration of this solution in PPM?
(use a formula!!!)
61. You have a very weak salt water solution that contains
just 1502 grams NaCl dissolved into 312,000 liters of
water. What is the molarity of this solution?
grams
solute
PPM =
x 1,000,000 =
grams solution
PPM =
1502 g NaCl
312,000,000 g solution
x 1,000,000 =
PPM = 4.81 parts per million NaCl in solution
(3 SF)
One tiny CoBr2 crystal has a mass of 0.00600 grams. If you dissolve it into
water to form a one liter solution of CoBr2(AQ), what is the molarity of this,
what is the solution’s concentration in PPM? (2 formulas first)
62.
63.
M=
PPM =
One tiny CoBr2 crystal has a mass of 0.00600 grams. If
you dissolve it into water to form a one liter solution of
CoBr2(AQ), what is the molarity of this, what is the
solution’s concentration in PPM? (2 formulas first)
M=
moles CoBr2
liters solution
M = 0.0000274 moles CoBr2
1.00 Liters
M = 0.0000274 M
or
274 ten millionths molar!
(3 SF)
PPM =
grams CoBr2
x 1,000,000
grams solution
0.00600 g CoBr2
PPM = 1000 g solution
PPM = 6.00 PPM
x 1,000,000
(3 SF)
This is a more “normal” number for
the normal brains we have.
There are 3 properties of water that are called the colligative properties.
They are the properties that are affected by dissolving solutes into the
water. Other solutions will have colligative properties but not in our intro class.
The 3 colligative properties of water are:
64. ________________________________
65. ________________________________
66. ________________________________
The reason for these properties are (remember, if it’s water’s properties,
there’s only 1 reason!)
the reason for the above 3 properties is because water has
67. _________________________ ___________________
68. Water has a normal boiling point of _____________________
69. Water has a normal freezing point of _______________________
70. Water has normal vapor pressure of ___________________________
(how did you know this?)
71. Water boils when it all has enough energy to break up all of the
____________________ ___________ that hold it together as a liquid.
72. If we add NaCl, this salt will ____________________ or it will
____________________________ in the water.
73. In solution the water molecules are hydrogen bonded to each other,
and they will also be __________________________ to the ions.
74. This will create more __________________ ___________________
which will ____________________ the boiling point.
75. Mathematically, the
BOILING POINT
_______________
for water is:
___________________________________________.
It is very important to see that
76. One mole of a molecular compound dissolved into a liter of water* has
______ of particles in solution.
77. One mole NaCl dissolved into a liter of water* has ______ of particles
in solution.
Crazy now:
78. One mole of CaCl2 into water dissolved into a liter of water*
has ______ of particles in solution.
Real Crazy now:
79. One mole of AlCl3 into water dissolved into a liter of water*
has ______ of particles in solution.
Watch your ionic compounds as they ionize into solution and COUNT the total
number of ions!
80. How many moles of
particles does 2
moles of AgCl form in
one liter of water?
_________
And don’t you forget this!
81. Calculate the temperature that a 1.00 liter,
2.00 M NaCl(AQ) solution will boil in Kelvin
temperature. Remember, each mole of particles
will elevate the BP by _____. So,
81.
Calculate the temperature that a 1.00 liter, 2.00 M NaCl(AQ) solution will
boil in Kelvin temperature.
Remember, each mole of particles will elevate the BP by 0.50 K. So,
Start BP
+
BP Elevation
= New BP
373 K
+
(4 x 0.50 K)
= 375 K
82. Calculate the Kelvin BP of a 1.00 Liter,
3.00 M CaCl2(AQ).
82. Calculate the Kelvin BP of a 1.00 Liter, 3.00 M CaCl2(AQ) .
Start BP +
BP Elevation = New BP
373 K + (9 x 0.50 K)
= 377.5 K
Each mole of calcium chloride yields 3 moles of ions,
3 moles of calcium chloride yields 9 moles of ions.
Each mole of ions elevates the BP by 0.50 K
The New BP will be equal to the original BP for water + the BP elevation
373 K + (9 x 0.50 K) = 373 K + 4.5 K = 377.5 K
(the heck with sf!)
83. The freezing point is _____________________ when you put particles
into solution. That’s because they “get in the way” of the water molecules
forming into their neat hexagons.
84. The FREEZING POINT __________________
for water is: ___________________________.
85. For every mole of particles (per liter of solution) the FP drops
__________________ K (or ______________°C)
 The ΔT°C = ΔT Kelvin
86. Calculate the temperature that a 1.00 liter,
2.00 M NaCl(AQ) solution will freeze in Kelvin.
86. Calculate the temperature that a 1.00 liter, 2.00 M NaCl(AQ) solution
will freeze in Kelvin.
Start FP
-
FP Depression
= New FP
273 K
-
(4 x 1.86 K)
= 266 K
Each mole of particles will depress the freezing point by 1.86 K, so
the FP changes to:
273 – (4 x 1.86 K) = 273 k – 7.44 K = 266 K
(with 3 SF)
87. Calculate the FP in Kelvin of a 1.00 Liter,
3.00 M CaCl2(AQ) .
87. Calculate the FP in Kelvin of a 1.00 Liter, 3.00 M CaCl2(AQ) .
Start FP
-
FP Depression
= New FP
273 K
-
(9 x 1.86 K)
= 256 K
Each mole of calcium chloride yields 3 moles of ions,
3 moles of calcium chloride yields 9 moles of ions.
Each mole of ions depresses the FP by 1.86 K
The New BP will be equal to the original FP for water - the FP depression
273 K + (9 x 1.86 K) = 273 K – 16.7 K = 256 K (3 SF)
88. Express the concentration of the following solution
in parts per million: 98.0 g of lithium chromate
(Li2CrO4) is dissolved into an aqueous solution with
total volume of 57,800 liters.
PPM =
88.
Express the concentration of the following solution in
parts per million:
98.0 g of lithium chromate (Li2CrO4) is dissolved into
an aqueous solution with total volume of 57,800
liters.
PPM =
98.0 g
X 1,000,000 = 1.70 PPM
57,800,000 g
89. We would use molarity when the solution has a
concentration that is _______________________
90. We’d use PPM when the concentration of solution is
______________________ __________________.
91. Vapor Pressure means
_____________________________________________________________
_____________________________________________________________
92. and it has units of ________________________ and you can find it on
__________________ ____
93. We won’t be doing any math with vapor pressure in HS chem, but
when solutions have lots of ions in solution, the VP will be
___________________ because _________________________________
______________________________________________________________
__________________________.
We can rank the vapor pressure of these solutions from lowest to highest by
comparing the number of moles of particles per liter in each. Rank these
1.0 liter aqueous solutions by vapor pressure.
Aqueous solution
94
1.00 M NaCl
95
1.00 M CaCl2
96
1.00 M NBr3
97
1.00 M Al(ClO3)3
Number of moles of ions per liter
Lowest VP Rank
We can rank the vapor pressure of these solutions from lowest to highest by
comparing the number of moles of particles per liter in each. Rank these
1.0 liter aqueous solutions by vapor pressure.
Aqueous solution
Number of moles of
particles per liter
Lowest VP
Rank
94
1.00 M NaCl
2
2nd highest
95
1.00 M CaCl2
3
3rd highest
96
1.00 M NBr3
1 (it’s not ionic!)
Highest VP
97
1.00 M Al(ClO3)3
4
Lowest VP
Next we will examine the dissociation of ionic compounds into water.
We’ll count ions!
Compound
98
Sodium carbonate
99
Ammonium sulfide
100
Aluminum nitrate
101
Lead (IV) acetate
102
Silver chloride
Formula
Ions formed when
put into water
Next we will examine the dissociation of ionic compounds into water.
We’ll count ions!
Compound
Formula
Ions formed when
put into water
98
Sodium carbonate
Na2CO3
Na+1 + Na+1 + CO3-2
99
Ammonium sulfide
(NH4)S2
NH4+1 + NH4+1 + S-2
100
Aluminum nitrate
Al(NO3) 3
Al+3 + THREE NO3-1
101
Lead (IV) acetate
Pb(C2H3O2)4
Pb+4 +
FOUR C2H3O2NO3-1
102
Silver chloride
AgCl
none ! It’s not AQ!
103. What is the freezing point of a 1.00 M
lead (IV) acetate solution of
1.00 liter volume? Don’t sweat SF
103.
What is the freezing point of a
1.00 M lead (IV) acetate solution
of 1.00 liter volume? Don’t sweat SF
1 mole Pb(C2H3O2)4(AQ) provides 5 moles of ions in solution
273 Kelvin – (5 x 1.86 K) = new FP
273 K – (9.3 K) = 263.7 Kelvin
104. President Obama calls upon you to help the Queen of
England who’s visiting America and has dined upon
3 dirty water hot dogs in the streets of Manhattan.
You must make her the antidote for the acid indigestion
she has developed. You need to make 151 mL of a 0.750 M
Ca(OH)2(AQ) for her to drink to neutralize that acid.
You only have 4.25 M calcium hydroxide(AQ) in stock!
Oh, she’s so fussy! How
can you help to avoid an
international incident and
make us (US?) look good?
(to the 10th mL, SF who cares!)
How do we mix up
this antidote!
104.
M1V1 = M2V2
(4.25 M)(V1) = (0.750 M)(151 mL)
V1 = 26.6 mL stock needed
151 mL total – 26.6 stock = 124.4 mL water
You get a Medal with a Star!
105. You find a solution in the back of the lab labeled
2.46 M KCl(AQ). It is filled to a line marked 2.00 Liters.
How many grams of KCl are in fact in this flask if the
label is correct?
105. You find a solution in the back of the lab labeled
2.46 M KCl(AQ). It is filled to a line marked 2.00 Liters.
How many grams of KCl are in fact in this flask if the
label is correct?
This label “means” 2.46 moles of KCl per liter. You have 2
liters, so this means 2 x 2.46 moles = 4.92 moles of KCl
the molar mass of KCl is (39 + 35 = ) 74 g/mole
so,
4.92 moles KCl
1
X
74 g KCl
1 mole KCl
= 364 g KCl
(3 SF)
106. According to an article in the New England Journal
of Medicine, Vol. 349, October 30, 2003,
mercury toxicity begins at 0.100 PPM.
If a crazy person dropped one pound of mercury into a school
pool, with a volume of 1,129,000 liters, would you be able to
safely swim in there? (1 pound = 454 grams)
According to an article in the New England Journal of Medicine, Vol. 349, October 30,
2003, mercury toxicity begins at 0.100 PPM.
106. If a nut dropped one pound of mercury into a school pool, with a
volume of 1,129,000 liters, would you be able to safely swim in there?
Short answer: NO.
PPM =
grams Hg
grams solution
PPM =
454 g Hg
1,129,000,000 g
PPM =
454 g Hg
1,129,000,000 g
PPM = 0.402 PPM
x 1,000,000
x 1,000,000
x 1,000,000
this is more than 4x the safe levels of mercury
107. What is the molarity of a solution
where 148 g KCl is dissolved into a
solution of 5000. mL total volume?
(you better start with a formula if you know what’s good for you!)
107. What is the molarity of a
solution where 148 g KCl is dissolved
into a solution of 5000. mL total
volume?
M=
moles of solute
liters of solution
2.00 moles KCl
M=
5.000 liters
= 0.400 molar
(3sf)
108. How do you mix up a 25.5 mL solution
that is 0.850 Molar NaOH(AQ), if you use
a stock solution of 6.40 M?
Then draw a diagram to show how to “make” this solution.
108.
How do you mix up a 25.5 mL solution that
is 0.850 Molar NaOH(AQ), if you use a stock
solution of 6.40 molarity. Then draw a
diagram to show how to “make” this solution.
M1V1 = M2V2
(6.40 M)(V1) = (O.850 M)(25.5 mL)
V1 = 3.39 mL stock
(3 sf)
Next slide please… this is NOT it.
25.5 mL total volume – 3.39 mL stock = 22.1 mL water for the dilution (3 SF)
That’s the 25.5 mL fill up
to here line
SECOND: fill to 25.5 mL total volume
(with water)
FIRST: put 3.39 mL stock solution into beaker.
109. You dissolve 2.25 moles of KBr into water forming
a 1.00 liter solution. What is this solution’s boiling
point, and freezing point, IN KELVIN?
(don’t worry about the SF in this problem, please)
109. You dissolve 2.25 moles of KBr into water forming a 1.00 L
solution. What is this solution’s BP and FP IN KELVIN?
(don’t worry about the SF in this problem, please)
Since KBr ionizes into 2 moles per mole of KBr,
2.25 moles KBr forms 4.50 moles of ions in total.
The original FP was 273 Kelvin, so the NEW FP is…
273 – (4.5 x 1.86K) = 273 – 8.37 = 264.63 K
The original BP was 373 Kelvin, so the NEW BP is…
373 + (4.5 x 0.50K) = 373 + 2.25 = 375.25 K
110. You have two 1.0 liter glasses of solution of equal
volume in the same room (same temp and pressure).
One solution is a 3.50 M NaCl aqueous solution, the
other is a 3.00 M Ca(NO3)2(AQ)
Which one would evaporate dry first, and why?
You must “prove” your answer with some math
110.
You have two 1.0 liter glasses of solution of equal volume in the same room
(same temp and pressure). One solution is a 3.50 M NaCl aqueous solution, the
other is a 3.00 M Ca(NO3)2(AQ) Which one would evaporate dry first, and why?
You must “prove” your answer with some math
3.50 M NaCl contains 7.0 moles of ions.
The CaCl2 has (3 x 3 = ) 9.0 moles of ions, so
The CaCl2 has greater internal attraction, and therefore a
lower vapor pressure.
The sodium chloride solution should evaporate faster.
111. If you have a 2.40 M HCl stock solution, how do
you make a 50.0 mL of 3.00 M HCl solution from it?
A diagram might help you think through this math.
111. If you have a 2.40 M HCl stock solution, how do
you make a 50.0 mL of 3.00 M HCl solution from it?
A diagram might help you think through this math.
M1V1 = M2V2
(2.40 M)(V1) = (3.00 M)(50.0 mL)
V1 = 62.5 mL stock ???? What???
Next slide please… this is NOT it.
111.
V1 = 62.5 mL stock
WAIT:
If you put 62.5 mL of stock in
how can the total solution be
just 50.0 mL???
It can’t: you can’t make this
happen, YOU CAN ONLY MAKE
MORE DILUTE SOLUTIONS!
FIRST: put 62.5 mL stock HCl.
(3 sf)
112. What volume of solution contains
475 g of sodium chloride at 0.933 Molarity?
112. What volume of solution contains
475 g of sodium chloride at 0.933 Molarity?
Moles solute
Molarity = Liters of Solution
475 g NaCl
1
X
1 mole NaCl
58 g NaCl
= 8.19 moles NaCl
8.19 Moles solute
0.933 M = Liters of Solution
(0.933)(Liters) = 8.19
Liters =
8.19 = 8.78 liters
0.933
114. Here are 3 labeled 1.0 Liter solutions.
Which has the highest boiling point, which has
the lowest freezing point?
3.00 M NCl3(AQ)
2.00 M NaCl (AQ)
1.50 M CaCl2(AQ)
114. Here are 3 labeled 1.0 Liter solutions.
Which has the highest boiling point, which has
the lowest freezing point?
3.00 M NCl3(AQ)
3 x 1 = 3 moles particles
2.00 M NaCl (AQ)
2 x 2 = 4 moles ions
1.50 M CaCl2(AQ)
1.50 x 3 = 4.5 moles ions
115. When 3.50 moles of zinc nitrate
become aqueous in water. In a one
liter solution of this, what is the boiling
point elevation, and the freezing point
depression?
(In Kelvin. Do not tell me the new BP
or new FP, I want the ΔT)
115. 3.50 moles of zinc nitrate become aqueous in water. In a one liter
solution of this, what is the boiling point elevation, and the
freezing point depression?
(In Kelvin. Do not tell me the new BP or new FP, I want the ΔT)
Zn(NO3)2 is the formula.
One mole of zinc nitrate forms three moles of ions.
3.50 x 3 = 10.5 moles of ions in solution. So,
BP elevation is 10.5 x 0.50K = 5.25 Kelvin increase
FP depression is 10.5 x 1.86 K = 19.5 K decrease
116. You have the misfortune of letting 8.30 g of H2O(G)
condense onto your arm. It cools rapidly to
55.0 °C. What is the total amount of energy that
was released (mostly into your arm)? (formula, or else!)
116. You have the misfortune of letting 8.30 g of H2O(G) condense onto
your arm. It cools rapidly to 55.0 °C. What is the total amount of energy
that was released (mostly into your arm)? (formula, or else!)
Condensation first,
q = mHV
q = (8.30 g)(2260 J/g)
q = 18,758 = 18,800 Joules
Then cool it down some
q = mCΔT
q = (8.30 g)(4.18 J/g·K)(45.0 K)
q = 1561.23 = 1560 Joules
Finally, sum the total joules together: 18,800 + 1560 = 20,360
answer: 20,400 Joules
117. Three ions are shown below. Draw in 3 water molecules properly oriented
to each of these three ions.
Cl-1
Cl-1
Ca+2
117. Three ions are shown below. Draw in 3 water molecules properly oriented
to each of these three ions.
O
O
Cl-1
H
O
H
Cl-1
H
H
H
H
O
O
Ca+2
O
O
H
H
118.
You prepare a 235 mL saturated solution of ammonium
chloride at 20.0°C. You go to lunch and come back in an hour.
The room temperature has warmed up this solution by about
five degrees centigrade. How would you best describe this
solution at 25.0 centigrade?
Be able to explain your choice.
A. Saturated at 25.0°C
C. Unsaturated at 25.0°C
B. Supersaturated at 25.0°C
D. Still saturated at 20.0°C
118.
You prepare a 235 mL saturated solution of ammonium chloride at 20.0°C.
You go to lunch and come back in an hour. The room temperature has
warmed up this solution by about five degrees centigrade. How would
you best describe this solution at 25.0 centigrade?
Be able to explain your choice.
C. Unsaturated at 25.0°C
The solution becomes unsaturated
because at 25.0°C this amount of
water can hold more NH4Cl than
when it was at a cooler temp.
119. If you have lots of sulfur solid floating on your
pond (or lots of water strider bugs), and you wanted
to clear the surface, you could add some soap.
Explain in a decent English sentence how this works.
119. If you have lots of sulfur solid floating on your pond (or lots of water
strider bugs), and you wanted to clear the surface, you could add some
soap. Explain in a decent English sentence how this works.
Surfactants are polar molecules that disrupt the
hydrogen bonding by interfering with this bonding
by getting in the way of the water molecules at
the surface.
Once the bugs/sulfur
gets through the surface,
down to the bottom that
will sink.
120. Oil floats on water. Explain why it floats,
why it does not sink, and
more importantly, why it does not mix.
120.
Oil floats on water. Explain why it floats, why it does not sink, and
more importantly, why it does not mix.
Oil is a nonpolar molecule, and since like
dissolves like (water is polar) oil and
water will not mix together, they are
immiscible. is less dense than water, so
it would float, not sink.
121. Using a 1.00 M stock of sugar water,
tell how to make up a 26.0 mL solution
of 0.350 M sugar water?
121. Using a 1.00 M stock of sugar water, tell how to make
up a 26.0 mL solution of 0.350 M sugar water?
M1V1 = M2V2
(1.00 M)(V1) = (0.350 M)(26.0 mL)
V1 = 9.10 mL stock solution plus…
26.0 mL – 9.10 mL = 16.9 mL water