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Solutions Examples of all you need to know

Solutions Examples of all you need to know. Each problem type is different. If you make an error, those are the problems to ask me about, and we’ll do more just like the one you were confused with.

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Solutions Examples of all you need to know

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  1. SolutionsExamples of all you need to know Each problem type is different. If you make an error, those are the problems to ask me about, and we’ll do more just like the one you were confused with.

  2. 1. What is the Molarity of a solution containing 2.50 moles of solute dissolved into 3.70 liters of water?Almost without fail, write the molarity formula every time you see the word moles with a solution, or anytime you see a capital letter M. # moles of solute Liters of solution M = Do this math before clicking ahead.

  3. 1. What is the Molarity of a solution containing 2.50 moles of solute dissolved into 3.70 liters of water? # moles of solute Liters of solution M = 2.50 moles 3.70 Liters M = M = 0.676 molar

  4. 2. How many moles of MgCl2 are in 8.20 liters of 1.75 M solution? # moles of solute Liters of solution M = Do this math before clicking ahead.

  5. 2. How many moles of MgCl2 are in 8.20 liters of 1.75 M solution? # moles of solute Liters of solution M = # moles of solute 8.20 Liters 1.75 M = X = 14.4 moles MgCl2 with 3 SF

  6. 3. How many liters of water are needed to make a 1.45 molar solution using 1.95 moles of NaCl? # moles of solute Liters of solution M = Do this math before clicking ahead.

  7. 3. How many liters of water are needed to make a 1.45 molar solution using 1.95 moles of NaCl? # moles of solute Liters of solution M = 1.95 moles NaCl X Liters 1.45M = 1.45 x = 1.95 x = 1.34 Liters water

  8. Rate of Solvation or Rate of solute dissolving into solution. Name + describe the three ways to dissolve a cube of sugar, or a hunk of rock salt into some water. What are the 2 factors that most affect how much solute that can dissolve into a solvent. What is table G for?

  9. Rate of Solvation or Rate of solute dissolving into solution. Name + describe the three ways to dissolve a cube of sugar, or a hunk of rock salt into some water. 1. stir or agitate the solution, adding kinetic energy which increases the rate of dissolving. 2. heat the water, which adds kinetic energy 3. crush the solute into small pieces which adds surface area and will increase the solvation of the solute

  10. What are the 2 factors that most affect how much solute that can dissolve into a solvent.1. Temperature: usually hotter means that more solute will dissolve into a given volume of water, but sometimes the reverse is true2. increased pressure will affect only gas solutes, and will increase the amount that can dissolve into the solvent What is table G for?It shows how much of 10 solutes will dissolve into 100 mL of water, at any temperature. It shows what happens when temp changes on a solution, and allows you to do proportion math to determine grams of solute in any sized solution

  11. Using Stock Solutions to prepare new solutions (make a drawing, it will help you) Concentrated solutions are used to prepare less strong solutions all the time in the chemistry lab back room. We’ll do this in lab this week too. Before you pour, you do some simple math, to “KNOW” how much of your stock solution to pour, and how much more deionized water to fill your solution up to. Silly as it might sound, it’s important to note, you cannot make a stronger solution from a weaker one with water. You can’t “dilute” to make it “stronger”. In our course, we’ll only learn to make weaker solutions from stronger ones.

  12. How do you prepare a 235 mL solution of 0.950 M NaCl(AQ)from a 2.00 M stock solution? The formula for dilution of solutions is: Molarity x Volume=Molarity x Volume STOCK SOLUTION = NEW SOLUTION Which is abbreviated as: (M1)(V1)=(M2)(V2)

  13. How do you prepare a 235 mL solution of 0.950 M NaCl(AQ)from a 2.00 M stock solution? (M1)(V1)=(M2)(V2) (2.00)(V1)= (0.950)(235 mL) V1= 112 mL stock solution (3SF) 235 mL total solution mark 235 mL – 112 stock = 123 mL water needed to fill 112 mL stock solution

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  15. How do you prepare exactly 100.0 mL of 1.00 M HCl(AQ) from a concentrated 12.0 M stock hydrochloric acid solution? Write the formula, fill in the blanks, solve for the stock volume, draw a picture, do the subtraction, get points on the regents, smile, grow up and be happy.

  16. How do you prepare exactly 100.0 mL of 1.00 M HCl(AQ) from a concentrated 12.0 M stock hydrochloric acid solution? (M1)(V1)=(M2)(V2) (12.0)(V1) = (1.00)(100.0 mL) V1 = 8.33 mL 3 SF 100.0 mL 91.7 mL water to fill up to the 100.0 mL line 8.33 mL stock

  17. How do you prepare a 50.0 mL sodium hydroxide solution with1.95 M using a stock solution of 1.25 molarity?

  18. How do you prepare a 50.0 mL sodium hydroxide solution with 1.95 M using a stock solution of 1.25 molarity? You don’t, or, you can’t. Stock solutions are always stronger, or more concentrated than their “daughter solutions”.Did I trick you?

  19. Concentration is the same as saying solution strength. How much stuff is dissolved into the solution? Usually in chemistry it’s expressed with molarity, the number of moles of solute dissolved per liter of solution. This is always okay and correct, but sometimes when the concentration of solute is very, very low, the molarity is a hard to deal with decimal number. When your concentration is so low that molarity is weirdly low, use a new concentration concept called PARTS PER MILLION. The formula is on your reference tables, and it is: Parts Per Million = Grams of solute Grams of solution X 1,000,000

  20. One particular grain of table salt has a mass of 0.00600 grams. It’s dissolved into 1.00 liters of pure water. What is the concentration of this solution? # moles of solute 0.00103 moles NaCl Liters of solution 1.00 Liters water M = M = 0.000103 M (? What?) Sometimes numbers like this make very little sense, even though they are correct. When molarity is silly tiny like this, use PPM instead.

  21. One particular grain of table salt has a mass of 0.00600 grams. It’s dissolved into 1.00 liters of pure water. What is the concentration of this solution? Grams of solute Grams of solution PPM = X 1,000,000 0.00600 g 1,000 g X 1,000,000 = 6.00 parts per million or 6.00 PPM PPM =

  22. A practical joke gone bad, some dopey kid puts exactly one pound (454.0 g) of mercury into the school pool, which has been just filled to 350,500 liters this morning. What is the concentration of Hg in H2O in PPM? It’s best to write the formula, because, Paper is cheap, Knowledge is valuable.

  23. Grams of solute Grams of solution PPM = X 1,000,000 Quick heads up: one liter = 1000 mL = 1000 g 350,500 Liters = 350,500,000 mL = 350,500,000 g of water 454.0 g Hg 350,500,000 g water PPM = X 1,000,000 PPM = 1.295 PPM with 4SF

  24. When a single cube of sugar (mass = 25.3 g) is dissolved into a Jacuzzi tub containing 1150 liters of pure water, the water gets “sugary”. What is the concentration of sugar in this solution? Start with the formula, a smile, think, get your calculator, use units always. Watch out for the three significant figures!

  25. When a single cube of sugar (mass = 25.3 g) is dissolved into a Jacuzzi tub containing 1150 liters of pure water, the water gets “sugary”. What is the concentration of sugar in this solution? Grams of solute Grams of solution PPM = X 1,000,000 25.3 grams sugar 1,500,000 grams water PPM = X 1,000,000 This pool contains 22.0 PPM sugar.

  26. Colligative Properties Solutions and water (and other liquids) have colligative properties, which are affected by the amount of solute dissolved into them. The colligative properties are boiling point, freezing point, and vapor pressure (which is a measure of how well it evaporates). We’ll deal mostly with water’s colligative properties in our class. Anything that disrupts the hydrogen bonding between the water molecules will disrupt the colligative properties. The polar water molecules are “sticky” to each other, because they are polar (no radial symmetry) and the semi permanent dipoles that are created by the strongly polar bonds, makes this so. Polar molecules and ionic compounds that dissolve into water will disrupt the hydrogen bonding between the molecules, and therefore disrupt the colligative properties too. The math is easy, and the affects are both numeric and predictable by you.

  27. BOILING POINT ELEVATION… the boiling point of pure water is of course 100°C at standard pressure. That is because it takes that much kinetic energy as measured in temperature to break apart all of the hydrogen bonds that water has holding it together (plus overcoming the air pressure pressing down on the surface). The more particles dissolved into the water, the more internal attraction, the higher the temperature it will take to boil the solution. For every mole of particles in 1.0 liters of water, the boiling point is increased by 0.50°C . For example:

  28. BOILING POINT ELEVATION For every mole of particles in 1.0 liters of water, the boiling point is increased by 0.50°C . For example:

  29. FREEZING POINT DEPRESSION… for every mole of particles dissolved into 1.00 liters of pure water, the freezing point is depressed by 1.86°C. The water molecules (six at a time) get so cold, the kinetic energy is so low at the freezing point, that the kinetic energy they have to keep moving is not as strong as the hydrogen bonds between them, so they lock into 6 molecule rings (in 3D). This is ice forming. Polar molecles and ions block this from happening, and it requires a lower kinetic energy – lower temperature – to make the ice form. Only with this lower KE can the hydrogen bonds overcome the blocking of the ions or polar molecules dissolved into solution.

  30. Freezing Point Depression

  31. One harder colligative property problem. You dissolve 319 grams of sodium chloride into 2.00 liters of water and stir. What is the new boiling point, what is the new freezing point for this solution? First, let’s get the number of moles of sodium chloride, then the number of moles of particles that are dissolved. 319 g NaCl 1 1 mole NaCl 58 g NaCl 5.50 moles NaCl = x 2 moles of ions 1 mole NaCl 5.50 moles NaCl 1 = 11 moles of ions x

  32. You dissolve 319 grams of sodium chloride into 2.00 liters of water and stir. What is the new boiling point, what is the new freezing point for this solution? With 11.0 moles of ions now in solution (NaCl yields Na+1 + Cl-1) 11.0 moles 2.00 Liters 5.50 moles 1 Liters = This means that for each mole of particles per liter of water, the boiling point goes up by 0.50°C, or BP increases by 2.75°C to 102.75 = 103°C with 3 SF The freezing point depression is -1.86°C for every mole of particles per liter, so the same solution, with 5.50 moles per liter would have a freezing point of: 5.50 x 1.86°C LOWER than the zero centigrade pure water has. The freezing point depression is: 5.50 x 1.86°C = 10.23 = 10.2°C with 3 SF The freezing point for this solution would be -10.2°C.

  33. VAPOR PRESSURE Vapor pressure remains difficult to grasp for many of you. It’s on table H. Vapor pressure is the extra pressure that is exerted in a sealed container due to the evaporation of a liquid inside this system. The more that evaporates, the higher pressure this extra vapor pressure is exerting. The amount of evaporation is controlled by temperature of the liquid, and how much internal attraction this liquid has to itself. Table H shows 4 liquids including water, and shows this extra pressure amount that the liquids exert under a sealed system, and lots of temperatures. Look hard at the drawings on this next slide…

  34. 2 bell jars at room temperature, 25°C. The left is empty, right, contains water. The pressure in the jar at left is the same pressure that the room it’s has. At right, the pressure in the jar is the same as the room PLUS the 8 kPa, which is the vapor pressure for water at 25°C. Cup of water inside bell jar Which evaporates Table H shows that at 25°C the VP H2O = 8kPa

  35. Vapor Pressure In Regents Chem there’s no math concerning the changes in the vapor pressure colligative properties due to dissolving of particles into water. There is math, just not for us. The more particles that are dissolved, the more internal attraction the aqueous solution will have, therefore evaporation will be harder, the vapor pressure would be lower. The water is already hydrogen bonded to itself, with more and more particles dissolved, the water is also attracted to these polar molecules or ions in solution, therefore evaporation would only occur when the water could shake free of this extra attraction it would have to itself PLUS these particles. I’d read the SOLUTIONS DIARY and make a list of problems you don’t grasp and ask Mr. Arbuiso about them soon. Peace, love, and chemistry.

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