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**Entry Task: April 27th Friday**

Question: What is the molarity of 1.2 moles of calcium carbonate in 1.22 liters of water?

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**Agenda: Sign off Discuss Concentration ws Notes on Dilutions**

HW: Concentration and Dilution ws

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**500g water + 75 g ethanol= 575 g solution**

1. What is the percent concentration of 75.0 g of ethanol dissolved in g of water? 75.0 g ethanol = % NaHCO3 500g water + 75 g ethanol 500g water + 75 g ethanol= 575 g solution 75g NaHCO3 X 100 575 g solution = 13.0% ethanol solution

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**= 33.6% solution 14.2 g benzene = % solution**

2. What is the percent concentration of benzene in a solution containing 14.2 grams of benzene in 28.0 grams of carbon tetrachloride? 14.2 g benzene = % solution 28.0 g CCl g benzene 28.0 g CCl g benzene = 42.2 g solution 14.2g benzene X 100 42.2 g solution = 33.6% solution

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**= 54.3g NaOCl Xg NaOCl = 3.62% NaOCl 1500 g solution Xg NaHCO3 = 3.62**

3. You have grams of bleach solution. The percent by mass of the solute sodium hypochlorite, NaOCl, is 3.62%. How many grams of NaOCl are in the solution? Xg NaOCl = 3.62% NaOCl 1500 g solution Xg NaHCO3 = 100 1500 g solution 1500 X 3.2 = 5430 g 100 = 54.3g NaOCl

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**= X% by volume ethanol solution**

4. What is the percent by volume of ethanol in a solution that contains 35 ml of ethanol in 115 ml of water? 35 ml of ethanol = X% by volume ethanol solution 35 ml ethanol ml of water 35 ml ethanol X 100 150 ml solution = 23.3 %

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**= 25.0 % 25 ml of methanol = X% solution**

5. What is the percent by volume of methanol in a solution that contains 25 ml of methanol in 75 ml of water? 25 ml of methanol = X% solution 25 ml ethanol + 75 ml of water 25 ml ethanol X 100 100 ml solution = 25.0 %

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**= 0.148 M 40 g of C6H12O6 1 mole of C6H12O6 180.16 g C6H12O6**

6. What is the molarity of an aqueous solution containing 40.0g of glucose (C6H12O6) in 1.5L of solution? 40 g of C6H12O6 1 mole of C6H12O6 g C6H12O6 0.222 moles of C6H12O6 1.5 L solution = M

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**= 0.128 M 9.5 g of NaOCl 1 mole of NaOCl 74.44 g NaOCl**

7. What is the molarity of a bleach solution containing 9.5 g of NaOCl per liter of bleach? 9.5 g of NaOCl 1 mole of NaOCl 74.44 g NaOCl 0.128 moles of NaOCl 1.0 L solution = M

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**= 0.00814 M 1.55 g of KBr 1 mole of KBr 119.0 g KBr 0.013 moles of KBr**

8. Calculate the molarity of 1.60 L of a solution containing 1.55 g of dissolved KBr. 1.55 g of KBr 1 mole of KBr 119.0 g KBr 0.013 moles of KBr 1.60 L solution = M

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**X moles of Ba(OH)2 = 0.0500 M 0.125 L solution**

9. Calculate the number of moles of a 125 ml solution (convert to liters) M Ba(OH)2. X moles of Ba(OH)2 = M 0.125 L solution (0.125 L)(0.0500) = moles

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**89.9 g Ba(OH)2 X moles of Ba(OH)2 = 1.50 M 0.350 L solution**

10. Calculate the number of grams of a 350 ml solution (convert to liters) 1.50 M Ba(OH)2. X moles of Ba(OH)2 = 1.50 M 0.350 L solution (0.350 L)(1.5) = moles Ba(OH)2 0.525 mol of Ba(OH)2 171.3 g Ba(OH)2 1 mole of Ba(OH)2 89.9 g Ba(OH)2

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Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) after dilution (f) = MiVi MfVf =

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Dilutions: M1V1 = M2V2 Molarity and Volume of the original solution = Molarity and Volume of the diluted solution. Make sure the volumes match!!!

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Dilutions: M1V1 = M2V2 P1. What volume, in milliliters of 2.00M calcium chloride (CaCl2) stock solution would you use to make 0.50 L of M calcium chloride solution? Convert liters to milliliters. (2.00 M)( X) = (0.300 M )(500 ml) 150 = 75 ml 2.00M

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Dilutions: M1V1 = M2V2 P2. What volume of a 3.00M KI solution would you use to make L of a 1.25 M KI solution? (3.00 M)( X) = (1.25 M )(0.300L) 0.375 = L or 125 ml 3.00M

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Dilutions: M1V1 = M2V2 P3. If I add 25 ml of water to 125 ml of a 0.15 M NaOH solution, what will the molarity of the diluted solution be? (0.15 M)( 125 ml) = (X M )(150 ml) 18.75 = M 150 ml

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**Use the rest of the period to work on homework**

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II. Solution Concentration (p. 480 – 488)

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