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Solutions

Solutions. Molarity, Molality, Dilutions, Percent Solutions, & Mole Fractions. Concentration. Concentration is a value that represents the amount of solute dissolved in a solvent. Concentrated solutions have a large amount of solute relative to the solvent

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Solutions

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  1. Solutions Molarity, Molality, Dilutions, Percent Solutions, & Mole Fractions

  2. Concentration • Concentration is a value that represents the amount of solute dissolved in a solvent. • Concentrated solutions have a large amount of solute relative to the solvent • Dilute solutions have a small amount of solute relative to the solvent • Concentration can be measured in Molarity, molality, %solution, ppm (parts per million)

  3. How it is made… • 2 Molar Solution of NaCl • requires 2 moles of NaCl to be dissolved to make 1L of solution • 2 moles of salt (or 117 grams) are dissolved in 1L of water to make a 2M NaCl(aq) solution

  4. Molarity • Molarity is the moles of solute per liter of solution. • Note the volume is the total solution volume, not the volume of solvent alone. • M = moles = mass/molar mass Liters Liters • This is the most common way chemists measure concentration

  5. Practice Problem #1 • What is the molarity of HCl if 28g HCl is dissolved in 500.mL of solution? • mass=28.g molar mass of HCl= 36.46g/mol V= 500mL=0.500L (divide by 1000) M= moles = (mass/molar mass) Liter Liter = (28/36.46)moles 0.500L = 1.5 M

  6. Practice Problem #2 • How many grams of sodium nitrate (NaNO3 )are needed to make 2 liters of a 0.100M solution? M=0.100M m=? mm NaNO3 = 85g/mol 0.100M = (m/85)mols 2L (0.100M)(2L)(85)= m m=17g NaNO3

  7. Solving using Dimensional Analysis • Molarity can be used as a conversion factor. 2.0 L 0.100moles 85 grams = 17g 1 Liter 1 moles

  8. Example 3 • What is the molarity of a solution with 10.0 grams of AgNO3 is dissolved in 500.mL of solution. Molarity = Mass / Molar Mass Liters of Solution Must convert to Liters!! 500mL/ 1000 = 0.500 Liters M = 10.0g / 169.88 g/mol 0.500 Liters M = 0.118 Molar

  9. Example 4 • How many grams of KNO3 should be used to prepare 2.00 Liters of a 0.500M solution? Molarity = Mass / Molar Mass Liters of Solution 0.500 = mass / 101.11 2.0 (0.500)(2.0) = Mass 101.11 Mass = (0.500)(2.0)(101.11) Mass = 101.11 grams

  10. Example 5 • To what volume should 5.0 grams of KCl be diluted to in order to prepare a 0.25M solution? Molarity = Mass / Molar Mass Liters of Solution 0.25 = 5.0 / 74.55 Volume V= 0.268 Liters = 268 mL V(0.25) = 5.0/74.55 V = 0.0671 0.25

  11. Molality (m) • Molality is the concentration of a solution expressed in moles of solute per kilogram of solvent. moles of solute (mol) • Molality (m) = ----------------------------------- kilogram of solvent (kg)

  12. Molality Sample Problem • A solution was prepared by dissolving 17.1 g of glucose, C6H12O6, in 275 g of water. What is the molality (m) of this solution? 0.345 mol/kg or 0.345 m

  13. More molality examples

  14. Preparing Solutions from Other Solutions • Many times solutions are prepared by diluting concentrated solutions. These are called stock solutions. • Dilution = process of adding water to a stock solution to achieve the molarity desired for a particular solution. • Adding water increases the volume of the solution and this causes the concentration to decease. • Dilution Equation: M1V1 = M2V2

  15. Dilutions and Molarity • Use this formula to make a more dilute solution from a concentrated solution Molarity1Volume1 = Molarity2Volume2 (Concentrated) (Dilute) (Moles before) = (Moles after!) M1V1 = M2V2

  16. Sample Problem #1 How many milliliters of 1.0 Molar HCl are required to make 100 mL of 0.025 M HCl? M1V1 = M2V2 M1 = 1.0M V1 = ? M2 = 0.025M V2 = 100mL (1.0M)V1 = (0.025M)(100mL) V1 = (0.025M)(100mL) (1.0M) = 2.5 mL of 1.0M HCl required, then dilute to 100mL

  17. Sample Problem #2 • How much water should you add to the volume of 1.0M HCl you calculated above to make the solution? We need 2.5 mL of concentrated 1.0M HCl. Then we dilute to a final total volume of 100mL. Therefore, 100mL – 2.5mL = 97.5 mL of water should be added. **Water added = V2 – V1

  18. More dilution example problems

  19. Percent Solutions • Solutions can also be represented as percent of solute in a specific mass or volume of solvent. • For a solid dissolved in water, you use percent by mass. • % by mass = mass solute x 100 mass of solution *Mass of solution = solute mass + solvent mass

  20. For a liquid mixed with another liquid % by volume = volume solute x 100 Total volume of solution

  21. Example 1 What percent solution do you have if you dissolve 80 grams of NaCl in 1 liter of water? * 1 Liter = 1000mL = 1000grams for water % mass = 80 grams NaCl X 100 (80g + 1000g H2O) = 7.4 %

  22. Example #2 • What percent solution will you have if you mix 40mL of ethanol with 200 mL of water? • %volume = 40mLs x 100 (200mLs + 40mLs) = 16.7%

  23. What if the previous problem was worded as follows? • What percent solution will you have if you dilute 40mL of ethanol to a final volume of 200mL? % volume = 40mL x 100 200mL* = 20% *Denominator is always total volume. You must pay attention the wording in the problem!

  24. More Percent Solution Problems

  25. Mole Fraction (X) • Number of moles of a compound divided by the total number of moles of all species in the solution. • Adding all mole fractions should get you a value of 1 • Guess what the units are?

  26. Example • What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water? (molar mass of glycerol =92 g/mol)

  27. Answer:

  28. Example 2 • What is the mole fraction of NaOH in an aqueous solution that contains 22.8% NaOH by mass? (molar mass of NaOH is 40 g/mol)

  29. Answer:

  30. More Mole Fraction Examples

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