Bellwork Dec 4 – graded Take out a sheet of paper and answer…

1. Bellwork Dec 4 – gradedTake out a sheet of paper and answer… • What would be the Molarity if 3.5 moles of sodium chloride were dissolved in 1.75 Liters of water? • 2. What would the Molarity be if 0.75 Moles of glucose dissolved in 750 ml of water? (Something tricky is happening here) • 3. 80.5 grams of sodium chloride is dissolved in 3.5 Liters. Determine the Molarity. • 4. 125 grams of Calcium Fluoride is dissolved in 250 mL of water. Solve for the Molarity. 2M 1M 0.39 M 6.4M Crash course chemistry: Solutions

2. A solution is a homogenous mixture of 2 or more substances. The solute is(are) the substance(s) present in the smaller amount(s). What is being dissolved. The solvent is the substance present in the larger amount. What is doing the dissolving. 12.1

3. Concentration • As we make solutions, we care about how much solute is actually dissolved inside • Concentration is a measure of this! • We say that something is concentrated if there is a lot of it dissolved, or we say that something is dilute if there is not a lot dissolved.

4. Molarity moles solute ( M ) = Molarity total liters of solution Molarity is defined as the amount of moles of a compound dissolved in an amount of solvent (usually water). It can be solved with the equation:

6. Molarity – What is it? • A measure of how concentrated a solution is. • Think about making Kool-Aid. • What makes Kool-Aid moredelicious? • The more sugar you add, thesweeter the solution gets. • Adding more moles of sugar, without changing the volume makes the Molarity “Concentration” Increase

7. Practice: 1) What would be the Molarity if 3.5 moles of sodium chloride were dissolved in 1.75 Liters of water? 2) What would the Molarity be if 0.75 Moles of glucose dissolved in 750 ml of water? (Something tricky is happening here)

8. What to look out for! Molarity is defined as moles per LITER. So every calculation, you need to make sure you convert your volume to Liters. Additionally, Molarity is in Moles. If your starting amount is in grams, be sure to convert to moles.

9. Molarity Practice Round 2 1) 80.5 grams of sodium chloride is dissolved in 3.5 Liters. Determine the Molarity. 2) 125 grams of Calcium Fluoride is dissolved in 250 mL of water. Solve for the Molarity.

10. What mass of KI is required to make 500. mL of a 2.80 M KI solution? Changing the question a bit… 2.80 moles x 1L x 500mL = 1.4 moles L 1000mL 1.4 moles KI x 165.99 g KI = 232 g KI 1 mole KI 12.3

11. Mass of solute = % Mass (Mass solute + Mass solvent) % by Mass • Another way of referring to concentration • Basically % composition of a solution • Key info:The density of water is handy! Every 1mL = 1gram of H2O X 100

12. % Mass Practice • If 31 grams of ammonium bromide are dissolved to make a 287 gram solution, what is the % by mass of the solute? • If a 64 gram solution is 5.2% sodium acetate, how many grams of solute are inside? 11% 3.3g

13. % Mass Practice Round 2 • 1.940 grams of potassium nitrate are dissolved in 32.0 mL of water. What is the % Mass? • 0.243 moles of sodium chloride are dissolved into 250mL of water. What is the % Mass? 5.7% 5.4%

14. Dilution of Solutions using Molarity Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution by adding more solvent. When you add more solvent, what happens to the number of moles already present in the solution? Do they increase? Decrease? Stay the same?

15. M1 = Starting Molarity , V1 = Initial Volume of solution M2 = Final Molarity, V2 = Final (total) Volume of solution Since moles are constant before and after dilution, we can use the following formula for dilution calculations. M1V1=M2V2

16. Example Type 1 • If you take 75 mL of a 1.7 M CaO solution and add in an additional 75 mL of water, what is the new concentration? M1V1 = M2V2 (1.7M ) (75mL) = (x) (150mL) M2= 0.85 M Note that V2 is the TOTAL volume of the solution, not just how much water was added.

17. Example Type 2 • You need 800.mL of a 2.00 M H2SO4 solution, but only have 6.00 M stock solution. How many mL of your stock solution do you need to use for the dilution? M1V1 = M2V2 6.0M x V1 = 2.0M x 800.mL V1= 267mL So we need to start with 267mL of the concentrated acid… • How much water will need to be added? • We want a total of 800.mL so… 800. – 267 = 533mL of water!

18. How would you prepare 60.0 mL of 0.2 M HNO3 from a stock solution of 4.00 M HNO3? MfVf 0.200 x 60.0 Vi = = 4.00 Mi MiVi = MfVf Mi = 4.00 Vi = ? mL Mf = 0.200 Vf = 60.0 mL = 3 mL 3 mL of acid + 57 mL of water = 60 mL of solution 4.5

19. Solution Stoichiometry • When you mix solutions, there can be chemical reactions. Knowing the Molarity, or concentration, of these solutions is critical when determining theoretical calculations, percent yield, and limiting reactant.

20. Write the equation for this problem: Hydrochloric acid reacts with Calcium Hydroxide to produce… _HCl + _Ca(OH)2  Need to Balance the Coefficients: _CaCl2+_H2O

21. 2 HCl + 1 Ca(OH)2  1 CaCl2 +2 H2O • If 255 mL of a 1.75 M HCl solution was used, determine how many grams of calcium chloride would be produced:

22. _H3PO4 + _Zn  _Zn3(PO4)2 + _H2 • If 55.0 grams of zinc are going to completely react, how many Liters of 0.75 M Phosphoric acid solution are required?