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 solventFaster solvent, more dissolving action, faster juggling

Agitation or stirringIncreases kinetic energy, faster dissolving action

Excess surface area of solutemore 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 soluteLiters 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 NaClLiters of solution

= 0.500 moles NaCl0.6000 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 orit’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 of750. mL

M = # moles KClLiters of solution

= 2.00 Moles KCl0.750 Liters solution

This solution has a Molarity of 2.67, or

It’s a 2.67 M KCl(AQ)

25. How many grams of sodium nitrate are in a one liter aqueous solution that is saturated at 10°C?

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

water80 g

100 mLX g

1000 mL

100 X = 80,000 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

1x 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 =

34. What units go into this formula (only) – MOLES of solute and LITERS not mL! of solution

Moles of soluteLiters 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 KClLiters of solution

= 3.50 Moles KCl0.750 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 NaClwater

40 g100 mL

x g1000 mL

100 x = 40,000 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 NaCl1 X 1 mole NaCl

58 g NaCl = 6.90 moles NaCl

M = 6.90 moles1 liter

= 6.90 M NaCl(AQ)

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.

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 NaClLiters of solution

0.900M = 1

# moles NaCl2.50 Liters

Solve for # moles, hint: put Molarity over 1, then cross multiply

# moles = 2.25 moles NaCl is 58 g/mole, so, 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 = 1

# moles KOH3.20 Liters

# moles KOH = 3.84 moles KOH

3.84 moles KOH1

x 56 g KOH1 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…

Concentration 

Molarity =Moles of soluteLiters of solution

For mixing solutions from scratch, or measuring strength of solutions.

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)

(1.15 M)(1.64 L)2.50 M

 

V1 =

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)

x mL = (1.00 M)(135 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…

1. Put your 24.5 mL STOCK SOLUTION into a new beaker.

2. Fill up to exactly 135 mL with water. (135 – 24.5 = 110.5 mL water)

Stock solution

water

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 NaCl1

X 1 mole NaCl58 g NaCl

= 25.9 moles salt

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.

25.9 moles312,000 Liters

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?

PPM = x 1,000,000 =

grams solutegrams solution

PPM = x 1,000,000 =

1502 g NaCl312,000,000 g

solution

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 solutionPPM = grams CoBr2

grams solutionx 1,000,000

M = 0.0000274 moles CoBr2

1.00 Liters

M = 0.0000274 M

or 274 ten millionths molar!(3 SF)

PPM = 0.00600 g CoBr2

1000 g solutionx 1,000,000

PPM = 6.00 PPM (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

57,800,000 gX 1,000,000 = 1.70 PPM

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 Number of moles of ions per liter Lowest VP Rank

94 1.00 M NaCl    

95 1.00 M CaCl2    

96 1.00 M NBr3   

971.00 M

Al(ClO3)3

   

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 solutionNumber 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 Formula Ions formed when put into water

98 Sodium carbonate

99 Ammonium sulfide

100 Aluminum nitrate

101 Lead (IV) acetate

102 Silver chloride

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)4Pb+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

273 Kelvin – (5 x 1.86 K) = new FP

273 K – (9.3 K) = 263.7 Kelvin

1 mole Pb(C2H3O2)4(AQ) provides 5 moles of ions in solution

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/moleso,4.92 moles KCl

1 X 74 g KCl1 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 = x 1,000,000

PPM = x 1,000,000

PPM = x 1,000,000

PPM = 0.402 PPM this is more than 4x the safe levels of mercury

grams Hggrams solution

454 g Hg 1,129,000,000 g

454 g Hg 1,129,000,000 g

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 =

M = = 0.400 molar (3sf)

moles of soluteliters of solution

2.00 moles KCl5.000 liters

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.

FIRST: put 3.39 mL stock solution into beaker.

SECOND: fill to 25.5 mL total volume (with water)

That’s the 25.5 mL fill up to here line

25.5 mL total volume – 3.39 mL stock = 22.1 mL water for the dilution (3 SF)

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 (3 sf)

 

FIRST: put 62.5 mL stock HCl.

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! X

 

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? 

Molarity = Moles soluteLiters of Solution

0.933 M = 8.19 Moles soluteLiters of Solution

475 g NaCl 1 X 1 mole NaCl

58 g NaCl = 8.19 moles NaCl

(0.933)(Liters) = 8.19 Liters = = 8.78 liters 8.19 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) 2.00 M NaCl (AQ) 1.50 M CaCl2(AQ)

3 x 1 = 3 moles particles 2 x 2 = 4 moles ions 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. 

Cl-1

Cl-1

Ca+2

H H

O

H H H

H

O O

H H

O

H H

O

H

H

O

H H

O

H

H

O

H H

O

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 B. Supersaturated at 25.0°CC. Unsaturated 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