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CHAPTER 4

CHAPTER 4. First year . Solutions A By Dr. Hisham Ezzat 2011- 2012 . The Dissolution Process. Solutions are homogeneous mixtures of two or more substances. Dissolving medium is called the solvent . Dissolved species are called the solute .

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CHAPTER 4

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  1. CHAPTER 4 First year • Solutions A By Dr. HishamEzzat 2011- 2012

  2. The Dissolution Process • Solutions are homogeneous mixtures of two or more substances. • Dissolving medium is called the solvent. • Dissolved species are called the solute. • There are three states of matter (solid, liquid, and gas) which when mixed two at a time gives nine different kinds of mixtures. • Seven of the possibilities can be homogeneous. • Two of the possibilities must be heterogeneous.

  3. The Dissolution Process Seven Homogeneous Possibilities SoluteSolventExample • Solid Liquid salt water • Liquid Liquid mixed drinks • Gas Liquid carbonated beverages • Liquid Solid dental amalgams • Solid Solid alloys • Gas Solid metal pipes • Gas Gas air Two Heterogeneous Possibilities • Solid Gas dust in air • Liquid Gas clouds, fog

  4. Ways of Expressing Concentration • Qualitative Terms: • Dilute Solution – A dilute solution has a relatively small concentration of solute. • A concentrated solution has a relatively high concentration of solute.

  5. Quantitative Terms Quantitative expressions of concentration require specific information regarding such quantities as masses, moles, or liters of the solute, solvent, or solution The solution process: Polar materials dissolve only in polar solvents (NaCI/H2O), and non - polar substances are soluble in non - polar solvents. This is the first rule of solubility "like dissolves like" e.g: benzene in CCI4

  6. [A] Weight to Weight expression.

  7. 1. Weight percent (wt %) Number of grams of solute which present in 100 gram of solution.

  8. e.g. 10% by weight glucose means: • 10 gm glucose + 90 gm H2O = 100 gm solution. • % Solute = (10/100) x 100 = 10 %. • % Solvent = (90/100) x 100 = 90 %

  9. 2. Mole fraction (X):

  10. 3. Molality • Molality is a concentration unit based on the number of moles of solute per kilogram of solvent.

  11. N.B W solution = W solute + W solvent W solvent = W solution - W solute W solvent =dV solution - W solute Where d = density of the solution V = Volume of the solution

  12. Example 1: What is the molality of 12.5 % solution of glucose C6H1206, in water? M.wt. of glucose is 180.0 Solution: 1) in 12.5 % solution 12.5 gm C6H12O6 is dissolved in l00 gm solution. W solvent = 100 - 12.5 = 87.5 g H2O 2) no. of moles glucose = 12.5/180

  13. Example 2: What are the mole fractions of solute and solvent in a 1.0m aqueous solution? Solution: The molecular weight of H2O is 18.0 we find the number of moles of water in 100 gm of H2O. no of moles of H2O A 1.0 aqueous solution contains n solute =1.0 mol The mole fractions are X solute X water =

  14. [B] Weight to Volume expression:

  15. Molarity

  16. Example 3: • How many grams of concentrated nitric acid solution should be used to prepare 250 ml of 2.0M HNO3? The concentrated acid is 70.0 % Solution: a) 70 gm HNO3 l00 gm solution Mass of pure HNO3 mass of HNO3 solution =

  17. b) If the density of the concentrated nitric acid solution is 1.42 g/ml. What volume should be used? M.wt. (HNO3) =63 ml cone. NHO3 = (45/1.42) = 31.7 ml cone. HNO3

  18. Example 4: An aqueous solution of acetic acid was prepared by dissolving 164.2 gm of acetic acid in 800 ml of the solution. If the density of the solution was 1.026 gm/ml. M. wt of acetic acid = 60 Calculate: a) The molar concentration of the solution b) The molality c) The mole fraction of both the solute and the solvent d) The mole % e) The weight %.

  19. Solution: a) b) d = 1.026g/ml V = 800 ml W solution = V x d = 800 x 1.026 = 820.8 gm W slvent = 820.8 - 164.2 = 656.6 gm

  20. c) no. of acetic acid moles = 164.2 / 60 = 2.737 mole no. of H2O moles = 656.6 / 18 = 36.44 mole Mole fraction of acetic acid = Mole fraction of H2O = d) mole % acetic acid = 0.0699 x 100 = 6.99 % mole % of H2O = 0.9299 x 100 = 92.99 % e) percentage weight of acetic acid = percentage weight of H2O=

  21. Try ? • Five grams of NaCl is dissolved in 25.0 g of H2O. What is the mole fraction of NaCl in the solution? (Answer =0.0580) • 2. What is the mole percent NaCl in the previous problem 1 (Answer = 5.80 mol %) • 3. Ten grams of ascorbic acid (vitamin C), C6H8O6, is dissolved in enough water to make 125 ml of solution. What is the molarity of the ascorbic acid? (Answer = 5.80 mol %) • 4. What is the molality of NaCl in the solution in the previous problem 1? (Answer = 3.42 m) • 5. What is the mass percent of NaCl in the solution in the previous problem 1? (Answer = 16.7 %)

  22. Try ?Example 1 • Five grams of NaCl is dissolved in 25.0 g of H2O. What is the mole fraction of NaCl in the solution? solution: The formula weight of NaCl is 58.44, so 5.00 g of NaCl is

  23. The molecular weight of water is 18.02; so 25.0 g of H2O is • In this solution, then, the mole fraction of NaCl is 0.0580. (The mole fraction of water is 1.0000 - 0.0580, or 0.9420.)

  24. Example 2 • What is the mole percent NaCl in the of Example 1 • mol % NaCI = XNaCl X 100 = 5.80 x 10-2 x 100 = 5.80% • The solution is 5.80 mol % NaCI and 94.20 mol % H2O

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