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O.A # 31

O.A # 31. Solution. O.A # 32. Concentration. O.A # 33. Miscible. O.A # 34. Immiscible. O.A # 35. Soluble. O.A # 36. Insoluble. Page # 19. Solution Concentration The concentration of a solution is a measure of how much solute is dissolved in a specific amount of solvent or solution.

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O.A # 31

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  1. O.A # 31 Solution

  2. O.A # 32 Concentration

  3. O.A # 33 Miscible

  4. O.A # 34 Immiscible

  5. O.A # 35 Soluble

  6. O.A # 36 Insoluble

  7. Page # 19 Solution Concentration The concentration of a solution is a measure of how much solute is dissolved in a specific amount of solvent or solution. • Solvent: substance that dissolves the solute • Solute: substance dissolved by the solvent

  8. Molality • Molality (m) is the ratio of moles of solute dissolved in one kilogram of solvent. • Molality (m)= moles of solute Kg of solvent

  9. Practice Problem What is the molality (m) of a solution containing 10.0g Na2SO4 dissolved in 255 g of water Step 1: convert grams of solute to moles 10.0 g x 1 mole = 0.0704 moles 142.1 g Step 2: convert solvent mass to Kg (g÷1000), If needed! 255 g x 1 Kg = 0.255 Kg 1000 g Step 3: solve for molality (m) Molality (m)= moles of solute Kg of solvent = 0.0704 moles = 0.255 Kg = 0.276 m = 0.276 moles/Kg

  10. Molarity • Molarity(M) is the number of moles of solute dissolved per liter of solution. Molarity is also known as molar concentration. • Molarity (M) = moles of solute Liters of solution

  11. Practice Problem What is the molarity of an aqueous solution containing 40.0g of glucose (C6H12O6) in 1500 mL solution? Step 1: convert solute grams to moles 0.222 moles 40.0 g x 1 mole = 180.0 g Step 2: Convert solution to Liters(mL÷1000), If needed! Step 3: Calculate molarity (M) M = moles of glucose = Liters of sln 0.222 moles = 1.5 Liters =0.15 M = 0.15 moles/Liter

  12. Page # - What are Solutions? Page 454 1) Why are oil and water immiscible? 2)How can you predict if a solute will dissolve in a solvent? (In other words explain the “like dissolves like” rule.) 3) What are the 3 factors that affect the rate of solvation? (Collisions between particles) 4) How does Temperature affect the solubility of some substances? 5) How does pressure affect the solubility of some substances?

  13. Answers 1) Because there is little attraction between the polar water molecules and the non polar oil molecules 2)To determine whether a solvent and solute are alike, you must examine the bonding and polarity of the particles and the intermolecular forces between the particles. 3) a. Agitating the mixture b. Increasing surface area of the solute • Increasing temperature of the solvent 4) Increasing temperature increases solubility and vice versa 5) Increasing pressure increases solubility and vice versa

  14. Van Hoff’s factor ( i ): Number of particles a compound dissociates in solution for covalent molecules, i= 1 ( nm + nm) 1. CaBr2 Ca+2+Br-1 2 i = 3 2. C12H22O11 covalent molecule i= 1 3. Sr(NO3)2 Sr +2 + NO3-1 2 i = 3 4. F2 covalent molecule i= 1

  15. Page 20- Colligative Properties Depend on the amount of moles of solvent Boiling Point Elevation:Solutions boil at a higher temperature than their solvents Tb(solution) = Tb(solvent) + Tb Tb = i x m x Kb Tb = Boiling point elevation i = Van Hoff’s factor m = molality = moles / Kg Kb = boiling point constant

  16. solute solvent Example: Calculate Bp of the solution: 100.0 g of C12H22O11 in 350 g of phenol. For phenol TBp = 84 ºC , Kb = 3.56 ºC/ m Step 1:Solve for molality = moles solute / Kg solvent 100.0 g x 1 mole = 342.0 0.2924 moles 350 g x 1 Kg = 0.35 Kg 1000 g m = 0.2924 = 0.35 0.84 m

  17. C12H22O11 covalent molecule i = 1 Step 2: Find i ( SOLUTE) Step 3: Calculate boiling point elevation Tb = i xm x Kb = 1 x 0.84 x3.56 = 3.0 ºC Step 4: Solve for boiling point of solution Tb(solution) = Tb(solvent) + Tb = 84 ºC + 3.0 ºC = 87 ºC

  18. Freezing Point Depression • Solutions freeze at a lower temperature than their solvents Tf Tf(solution) = Tf(solvent) + Tf = - i x m x Kf Example: Calculate Fp of the solution: 280.0 g of MgCl2 in 2500 g of water. For water Tfp = 0.0 ºC , Kf = 1.86 ºC/ m solute solvent Step 1:Solve for molality = moles solute / Kg solvent 280.0 g x 1 mole = 95.3 g 2.938 moles 2500 g x 1 Kg = 2.5 Kg 1000 g m = 2.938 = 2.5 1.2 m

  19. Step 2: Find i ( SOLUTE) MgCl2 i = 3 Mg +2 + 2 Cl-1 Step 3: Calculate freezing point depression Tf = - i xm x Kf - 6.7 ºC = - 3 x 1.2 x1.86 = Step 4:Solve for freezing point of solution Tf(solution) = Tf(solvent) + Tf = - 6.7 ºC = 0.0 ºC + (-6.7 ºC )

  20. Quick Write Explain the similarities and differences between a 1M solution of NaOH and a 1m solution of NaOH.

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