1. New Seating Chart! Please pick up a new unit packet AND your quiz folder. I will pass back quizzes during the QOTD. QOTD 4/28/14: (your notes from last Wednesday will be helpful) You dissolve 3 moles of NaCl in enough water to make a total of 12 L of NaCl (aq) solution… What is the solvent? 2. What is the solute? 3. Calculate the concentration (molarity)

2. QOTD Answer 1. Solvent = the part you have more of = water
(12 Liters is about 3 gallons of water, but 3 moles of NaCl will fit in the palm of your hand
in general, the LIQUID will be the solvent……
2. Solute = what you have less of
what is dissolved in the liquid

3. QOTD answer 3. Molarity: M = mol of solute L of solution
3 mol / 12 L = mol/L = 0.25 M

4. Learning Target
You should be able to define the terms concentrated and dilute and draw a diagram to support your explanation.
You should also be able to explain what happened in the gummy bear experiment.

5. Please turn to page 7!!!
Do the activity on p We will go over the questions 1-3 in about 5 minutes.
If you finish early, work on Making Sense Qs and Qs 1-16 on p

6. Answers to
Unit 6.1.2: BEARLY Alive!
These answers are available on my website.

7. Rank the bear size from 1-6
Solution
Solute
Observations
Rank the bear size from 1-6
Water --
1 (biggest)
0.1 M sugar
C12H22O11 2
0.5 M sugar 3
1.0 M sugar 4
2.0 M sugar 5
Corn syrup
sugars
6 (smallest)

8. 1. What does it mean when chemists say that a 1
1. What does it mean when chemists say that a 1.0 M solution is more concentrated than a 0.1 M solution? (Hint: Mention moles and liters in your answer)
there are more molecules dissolved in the same amount of solution (essentially the same amount of liquid)
Specifically, there is 1 whole mole dissolved in every 1 L compared to only 1/10 mole in every 1 L

9. The more concentrated the solution, the smaller the gummy bear.
2. How does the size of the gummy bear relate to the concentration of the solution it is in?
The more concentrated the solution, the smaller the gummy bear.

10. 3. The gummy bear itself can be thought of as a sugar solution
3. The gummy bear itself can be thought of as a sugar solution. What do you think the Molarity of sugar is in the gummy bear?
~3M?

11. Please do the making sense Qs on the bottom of page 8 & Qs 1-16 on pages 9-10!
We will go over these during the last 15 minutes of class.
If you finish early, let me know and I will give you the homework WS.

12. Making Sense Questions:
What do you think is happening in this activity? Write a paragraph explaining your ideas. Include your answers to the following questions:
What causes the size of the gummy bears to change when added to water?
Why are gummy bears in sugar solutions different sizes based on concentration?
The gummy bears change sizes when added to water because water molecules move from an area of low solute concentration to an area of high solute concentration. The more solute (more concentrated) there was in a solution surrounding the gummy bear, the more likely the water from the gummy bear would rush out. If there was less solute surrounding the gummy bear, the abundant water outside of the bear would flow into the gummy bear.

13. 112 g S x 1 mol S = 3.49 moles S 32.07 g S Determining Molarity:
Part 1: Moles Review (show your work!)
1. How many moles are in 112 g of sulfur?
112 g S x 1 mol S = moles S
32.07 g S

14. Determining Molarity:
Part 1: Moles Review (show your work!)
2. How many grams does 2.2 moles of NaOH weigh?
Na (22.99) + O (16) + H (1.008) = g/mol
2.2 mol x g = 88 g
1 mol

15. C (12.01) + 2 O (16x2) = 44.01 g/mol 11 g CO2 x 1 mol = 0.25 mol
Determining Molarity:
Part 1: Moles Review (show your work!)
3. How many moles are in 11 grams of carbon dioxide (CO2)?
C (12.01) + 2 O (16x2) = g/mol
11 g CO2 x 1 mol = mol
44.01 g

16. Part 2: Practice Calculating Molarity (ALWAYS Show work!)
What is the concentration (in moles/L) when . . .
moles of glucose is dissolved in water to make 4.0 Liters of solution?
2.0 moles = 0.5 M
4.0 Liters

17. Part 2: Practice Calculating Molarity (ALWAYS Show work!)
What is the concentration (in moles/L) when . . .
moles of sodium hydroxide are dissolved in enough water to make a 1.50 Liter solution?
3.75 moles = 2.5 M
1.5 Liters

18. 0.0014 mol = 0.02 M 0.065 Liters 65 mL x 1 L = 0.065 Liters 1000 mL
Part 2: Practice Calculating Molarity (ALWAYS Show work!)
What is the concentration (in moles/L) when . . .
moles of KNO3 is dissolved in 65 mL?
65 mL x 1 L = Liters
1000 mL
mol = M
0.065 Liters

19. Part 2: Practice Calculating Molarity (ALWAYS Show work!)
What is the concentration (in moles/L) when . . .
moles of KBr are dissolved into 400 mL of solution.
400 mL x 1 L = 0.4 L
1000 mL
0.8 mol = 2 M
0.4 Liters

20. 1.25 mol = 0.42 M 3.0 Liters 225 g C6H12O6 x 1 mol = 1.25 mol
Part 3: Using mass to calculate molarity:
What is the molarity of the following solutions?
8) 225 grams of glucose (C6H12O6) is dissolved to make 3.0 liters of solution?
6 C (12.01 x 6) + 12 H (1.008 x 12) + 6 O (16 x 6) = g/mol
225 g C6H12O6 x 1 mol = mol g
1.25 mol = M
3.0 Liters

21. 2 mol = 0.4 M 5.0 Liters 170 g NaNO3 x 1 mol = 2 mol 85 g
Part 3: Using mass to calculate molarity:
What is the molarity of the following solutions?
9) 170 grams of NaNO3 is dissolved to make 5.0 liters of solution?
Na (22.99) + N (14.01) + 3 O (16 x 3) = 85 g/mol
170 g NaNO3 x 1 mol = 2 mol
85 g
2 mol = M
5.0 Liters

22. 1.5 mol = 3 M 0.5 Liters 249 g KI x 1 mol = 1.5 mol 166 g
Part 3: Using mass to calculate molarity:
What is the molarity of the following solutions?
10) A 500 mL solution which contains 249 g of potassium iodide?
K (39.1) + I (126.9) = 166 g/mol
249 g KI x 1 mol = mol
166 g
1.5 mol = 3 M
0.5 Liters

23. M x L = moles 2.5 mol x 1.5 L = 3.75 moles of LiF 1 L
Part 4: Using molarity to solve problems.
11) How many moles of LiF would be required to produce a 2.5 M solution with a volume of 1.5 L?
M x L = moles
2.5 mol x 1.5 L = 3.75 moles of LiF
1 L

24. Moles ÷ M = Liters Part 4: Using molarity to solve problems.
12) How many liters of a 2.5 M solution can be made using 6 moles of sodium carbonate?
Moles ÷ M = Liters
6 moles ÷ 2.5 moles/1L = 6 mol x 1 L = L
2.5 mol

25. 3.5 M x 2.0 L = 7 moles CaCl2 Ca (40.08) + 2 Cl (35.45) = 110.98 g/mol
Part 4: Using molarity to solve problems.
13) How many grams of CaCl2 would be required to produce a 3.5 M solution with a volume of 2.0 L?
M x L = moles, moles x molar mass = grams
3.5 M x 2.0 L = 7 moles CaCl2
Ca (40.08) + 2 Cl (35.45) = g/mol
7 mol CaCl2 x g = g
1 mol

26. 0.99 mol = 0.2 Liters 5.0 mol/Liter 100 g KNO3 x 1 mol = 0.99 mol
Part 4: Using molarity to solve problems.
14) How many liters of a 5.0 M solution can be made using 100 grams of KNO3?
K (39.1) + N (14.01) + 3 O (16 x 3) = g/mol
100 g KNO3 x 1 mol = mol g
0.99 mol = Liters
5.0 mol/Liter

27. M x L = moles 3.5 moles x 2.50 L = 8.75 mol 1 L
Part 4: Using molarity to solve problems.
15) How many moles of Sr(NO3)2 would be used in the preparation of 2.50 L of a 3.5 M solution?
M x L = moles
3.5 moles x 2.50 L = 8.75 mol
1 L

28. M x L = mol , mol x g/mol = grams
Part 4: Using molarity to solve problems.
16) How many grams of NaI would be used to produce a 2.0 M solution with a volume of 500 mL?
M x L = mol , mol x g/mol = grams
2.0 mol x 0.5 L = 1 mol
1 L
Na (22.99) + I (126.9) = g/mol
1 moles NaI x g/mol = g

29. Exit Ticket!
Please answer the questions on the exit ticket. Be sure to put your name on it!
Homework = Molarity Practice Problems WS
Back side will be graded for correctness! The possible answers are at the bottom.