1. Molarity 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: Molarity (M) = moles solute liters of solution

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

3. 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)

4. 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.

5. 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.

6. 12.3 What mass of KI is required to make 500. mL of a 2.80 M KI solution?

7. Dilution of Solutions Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. When you want to dilute a solution, what happens to the number of moles present in the solution? –Do they increase? –Decrease? –Stay the same?

8. Since moles are constant before and after dilution, we can use the following formula for dilution calculations. M 1 V 1 =M 2 V 2 M 1 = Starting Molarity, V 1 = Initial Volume of solution M 2 = Final Molarity, V 2 = Final Volume of solution

9. Example Describe how you would prepare 800mL of a 2.0M H 2 SO 4 solution, starting with a 6.0 M stock solution. M 1 V 1 = M 2 V 2 6.0M x V 1 = 2.0M x 0.800L V 1 = 1.6M x L/ 6.0M V 1 = 0.26L How much water will need to be added to make this diluted solution?

10. How would you prepare 60.0 mL of 0.2 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M i V i = M f V f M i = 4.00 M f = 0.200V f = 0.06 L V i = ? L 4.5 V i = MfVfMfVf MiMi = 0.200 x 0.06 4.00 = 0.003 L = 3 mL 3 mL of acid + 57 mL of water= 60 mL of solution

11. Officially, C = Molarity, n = Moles, v = Volume of Solution

12. 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.

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

14. 2 HCl + 1 Ca(OH) 2  1 CaCl 2 +1 H 2 O If 255 mL of a 1.75 M HCl solution was used, determine how many grams of calcium chloride would be produced:

15. _H 3 PO 4 + _Zn  _Zn 3 (PO 4 ) 2 + _H 2 If 55.0 grams of zinc are going to completely react, how many Liters of 0.75 M Phosphoric acid solution are required?