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Solutions Chem Class #1

Solutions Chem Class #1. OB: Describing what solutions are, how they form, and how are they’re strength is measured.

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Solutions Chem Class #1

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  1. Solutions Chem Class #1 OB: Describing what solutions are, how they form, and how are they’re strength is measured.

  2. You cannot proceed unless you can discuss the differences between solute and solvent. You also need to know definitions and some examples of solutions that are unsaturated, saturated, and supersaturated. A solution is a homogeneous mixture when a solute is dissolved into a solvent. Strangely they are aqueous, dissolved into other liquids, or even solids (alloys) or gases (like air).

  3. What factors affect the rate of a solute dissolving into a solvent? 1. 2. 3.

  4. What factors affect the rate of a solute dissolving into a solvent? Higher temperature of solventFaster solvent, more dissolving action, faster juggling Agitation or stirringIncreases kinetic energy, faster dissolving action Excess surface area of solutemore surface on solvent allows more solvation, less “waiting” for a chance to swim in the solvent

  5. How much solute will dissolve into a solution? Depends first on actual solubility (table F) Temperature of solvent (usually hotter = more) Pressure (affects only gases) usually hotter temps = higher solubility

  6. Supersaturation… Not usual. When a solution is heated it can usually hold more solute. If the saturated solution is cooled slowly, without agitation, sometimes the solution can hold more solute than it could normally at the colder temperature. Sugar water can do this, and is done making rock candy. Sodium acetate does this, and is done so in re-usable hand warmers. Nothing on table G can supersaturate, that’s important to remember.

  7. High Pressure can only affect gases, not solid solutes, or liquids that are miscible. This is used a lot, especially when making soda.

  8. Measuring the strength of a solution. MOLARITY (M)Molarity is the expression of concentration of a solution as measured by the number of moles of solute in a liter of solution. M = # moles soluteLiters of solution

  9. Problem 1. What is the concentration of a one liter salt water solution containing 58 g NaCl? (don’t even start without the formula!)

  10. Problem 1. What is the concentration of a one liter salt water solution containing 58 g NaCl? (don’t even start without the formula!) 1.0 mole NaCl 1 Liter solution M = # moles NaClLiters of solution = This is a 1.0 Molar solution, or this solution is 1.0 M NaCl, or the solution contains 1 mole NaCl/Liter

  11. Problem 2. If you add 29.0 g NaCl to enough water to form a 600.0 mL solution, what is it’s concentration?(write a formula or else)

  12. Problem 2. If you add 29.0 g NaCl to enough water to form a 600.0 mL solution, what is it’s concentration?(write a formula or else) M = # moles NaClLiters of solution = 0.500 moles NaCl0.6000 Liters solution This solution is 0.833 M This solution has a molarity of 0.833 It’s am 0.833 M NaCl aqueous solution

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  14. Problem 3. You put 74.0 g KCl solid into a flask. You fill the flask to 1600.0 mL, what is the molarity of this solution?

  15. Problem 3. You put 74.0 g KCl solid into a flask. You fill the flask to 1600.0 mL, what is the molarity of this solution? # moles KClLiters of solution = M = 1.00 moles KCl1.6000 Liters solution This solution is 0.625 Molar KCl Or, this Potassium Chloride Aqueous Solution has a 0.625 Molarity

  16. Problem 4. Calculate the molarity of an aqueous solution containing 148 grams KCl in a750. mL solution.

  17. Problem 4. Calculate the molarity of an aqueous solution containing 148 grams KCl of750. mL M = # moles KClLiters of solution = 2.00 Moles KCl0.750 Liters solution This solution has a Molarity of 2.67, or It’s a 2.67 M KCl(AQ)

  18. Problem 5. How many grams of sodium nitrate are in a one liter aqueous solution that is saturated at 10°C?

  19. Problem 5. How many grams of sodium nitrate are in a one liter aqueous solution that is saturated at 10°C? NaNO3water 80 g100 mL x g1000 mL 10°C 100 x = 80,000 x = 800 grams NaNO3

  20. How many moles is 800. g NaNO3 ? Calculate the molarity of this solution. 800. g NaNO31 1 mole NaNO385 g NaNO3 x = 9.41 moles NaNO3 = 9.41 M 9.41 moles 1.0 liters

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