1 / 18

Entry Task: April 27 th Friday

Entry Task: April 27 th Friday. Question: What is the molarity of 1.2 moles of calcium carbonate in 1.22 liters of water?. Agenda:. Sign off Discuss Concentration ws Notes on Dilutions HW: Concentration and Dilution ws.

merlin
Télécharger la présentation

Entry Task: April 27 th Friday

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Entry Task: April 27thFriday Question: What is the molarity of 1.2 moles of calcium carbonate in 1.22 liters of water?

  2. Agenda: • Sign off Discuss Concentration ws • Notes on Dilutions • HW: Concentration and Dilution ws

  3. 1. What is the percent concentration of 75.0 g of ethanol dissolved in 500.0 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

  4. 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 CCl4+ 14.2 g benzene 28.0 g CCl4+ 14.2 g benzene = 42.2 g solution 14.2g benzene X 100 42.2 g solution = 33.6% solution

  5. 3. You have 1500.0 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? XgNaOCl = 3.62% NaOCl 1500 g solution Xg NaHCO3 = 3.62 100 1500 g solution 1500 X 3.2 = 5430 g 100 = 54.3g NaOCl

  6. 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 + 115 ml of water 35 ml ethanol X 100 150 ml solution = 23.3 %

  7. 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 %

  8. 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 180.16 g C6H12O6 0.222 moles of C6H12O6 1.5 L solution = 0.148 M

  9. 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 = 0.128 M

  10. 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 = 0.00814 M

  11. 9. Calculate the number of moles of a 125 ml solution (convert to liters) 0.0500M Ba(OH)2. X moles of Ba(OH)2 = 0.0500 M 0.125 L solution (0.125 L)(0.0500) = 0.00625 moles

  12. 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) = 0.525 moles Ba(OH)2 0.525 molof Ba(OH)2 171.3 g Ba(OH)2 1 mole of Ba(OH)2 89.9 g Ba(OH)2

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

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

  15. 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 0.300 M calcium chloride solution? Convert liters to milliliters. (2.00 M)( X) = (0.300 M )(500 ml) 150 = 75 ml 2.00M

  16. Dilutions: M1V1 = M2V2 • P2. What volume of a 3.00M KI solution would you use to make 0.300 L of a 1.25 M KI solution? (3.00 M)( X) = (1.25 M )(0.300L) 0.375 = 0.125 L or 125 ml 3.00M

  17. 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 = 0.125 M 150 ml

  18. Use the rest of the period to work on homework

More Related