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Solutions

Solutions. Solutions. A solution is made up of a solute dissolved in a solvent. Example: seawater is a solution. It is made up of salt (the solute) dissolved in water (the solvent). Solutions can be described as concentrated or dilute.

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Solutions

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  1. Solutions

  2. Solutions • A solution is made up of a solute dissolved in a solvent. • Example: seawater is a solution. • It is made up of salt (the solute) dissolved in water (the solvent).

  3. Solutions can be described as concentrated or dilute. • A concentrated solution has a large amount of solute per solvent • A dilute solution has a small amount of solute per solvent.

  4. Concentration of solutions • The concentration of a solution is the amount of solute dissolved in a certain amount of solvent.

  5. 1. Moles per litre Concentration can be measured in : • moles of litre (mol L-1 ). • This is also known as molarity. • 1M solution = • 1 mole in 1 litre = • 1 mole in 1000cm3

  6. 2. Grams per litre Concentration can be measured in : Grams per litre of litre (g l-1 ). • 1g l-1 solution = • 1g in 1 litre = • 1g in 1000cm3

  7. Example 1A solution contains 10g of Sodium Hydroxide in I litre of solution.What is the concentration expressed in moles per litre. How many moles in 1 litre of solution? 10g / RMM = moles 10g / 40 = 0.25 X Relative Molecular Mass Mass of X in g Moles of X ÷ Relative Molecular Mass There are 0.25 moles of NaOH in one litre. Answer = 0.25 mol L-1

  8. Example 2A solution contains 28g of Potassium Hydroxide (KOH) in I litre of solution.What is the concentration in moles per litre. X Relative Molecular Mass Mass of X in g Moles of X ÷ Relative Molecular Mass • 28g / RMM = moles • 28 / 56 = 0.5 • There are 0.5 moles of KOH in one litre. • Answer = 0.5 mol L-1

  9. Example 3:A solution contains 48g of MgSO4 in 3 litres of water.What is the concentration in moles per litre. Given Grams per 3 litres GRAMS PER LITRE Find MOLES PER LITRE • How many grams in 1L ? • 48/ 3 = 16 • There are 16g of MgSO4 in one litre.

  10. Given Grams per 3 litres GRAMS PER LITRE Find MOLES PER LITRE How many moles in 1 litre of solution? X Relative Molecular Mass Mass of X in g Moles of X ÷ Relative Molecular Mass • 16 / RMM = moles • 16 = 0.1333 • 120 • There are 0.1333 moles of MgSO4 in one litre. • Answer = 0.1333 mol L-1

  11. Q219 • Calculate the molarity of a solution which contains 65g of HCl in a litre X Relative Molecular Mass Mass of X in g Moles of X ÷ Relative Molecular Mass 65 / rmm = moles 65 / 36.5 = 1.7808 moles in a litre Answer : 1.7808M

  12. Q219b • Calculate the molarity of a solution which contains 65g of HCl in a litre X Relative Molecular Mass Mass of X in g Moles of X ÷ Relative Molecular Mass 25 / rmm = moles 25 / 56 = 1.7857 moles in 250cm3 1.7857 x 4 = 7.1428 moles in a litre Answer : 7.1428 M

  13. Q219c • Calculate the molarity of a solution which contains 22g of H2SO4 per 100cm3 of a solution X Relative Molecular Mass Mass of X in g Moles of X ÷ Relative Molecular Mass 22 / rmm = moles 22 / 98 = 0.2245 moles in 100cm3 0.2245 x 10 = 2.2448 moles in a litre Answer : 2.2448 M

  14. Q219d • Calculate the molarity of a solution which contains 10g of NaOH per 2L solution X Relative Molecular Mass Mass of X in g Moles of X ÷ Relative Molecular Mass 10 / rmm = moles 10 / 40 = 0.25 moles in 2L 0.25 /2 = 0.125 moles in a litre Answer : 0.125 M

  15. Q219e • Calculate the molarity of a solution which contains 12.5g of Na2CO3 per 200cm3 solution X Relative Molecular Mass Mass of X in g Moles of X ÷ Relative Molecular Mass 12.5 / rmm = moles 12.5 / 106 = 0.1179 moles in 200cm3 0.1179 x 5 = 0.5896 moles in a litre Answer : 0. 5896 M

  16. 221 • A chemist dissolves 98.4 g of FeSO4 in enough water to make up 2L of solution. • What is the molarity of the solution? X Relative Molecular Mass Mass of X in g Moles of X ÷ Relative Molecular Mass 98.4 / rmm = moles 98.4 / 152 = 0.6474 moles in 2L 0.6474 / 2 = 0.3237 moles in a litre Answer : 0. 3237 M

  17. Q220 • 0.080g of ethanol in 100cm3 of blood • 0.080 x 10 = 0.8g of ethanol in 1L of blood • 0.8/ RMM= Moles of ethanol • 0.8/ 46 = 0.0174 moles in 1L • Answer = 0.0174M

  18. Q223 • Calculate the mass of NaOH in 20cm3 of 0.1M NaOH X Relative Molecular Mass Mass of X in g Moles of X ÷ Relative Molecular Mass Moles x rmm = mass 0.1 x 40 = 4.0g in 1L 4.0g /1000 = .004g in 1cm3 0.004 x 20 = 0.08 g in 20cm3 Answer :0.08g

  19. q223 • B) 1.53g • C)0.0158g • D) 146g • E)0.196g

  20. Q224 • A) 40gl-1 • B)112gl-1 • C)12.25gL-1 • D)1.06gl-1 • E)0.49gl-1

  21. Example 4What is the concentration in grams per litre of a 0.025M solution of Ca(OH)2? Given Moles PER LITRE Find Grams PER LITRE • (0.025)(74) • There are 1.85g of Ca(OH)2 in one litre. • Answer = 1.85gL-1

  22. Example:What is the mass of CuSO4 in 250cm3 of a 3M solution? Given Moles PER LITRE Find Grams PER LITRE Grams per 250cm3 How many grams in 1 litre of solution? • There are 478.5g of CuSO4 in one litre. • Answer =478.5 gL-1

  23. Given Moles PER LITRE Find Grams PER LITRE Grams per 250cm3 How many grams in 250cm3 litre of solution? • (478.5)/ 4 = • Answer = 119.625grams in 1 litre • = 119.625gl-1

  24. 3. Mass per volume • Concentration is also measured in % mass of solute(g) per volume of solvent(cm3) • This is written as % (w/v) • 1% (w/v) means 1gof solute in 100cm3 of solvent • 20% (w/v) means 20g of solute in 100cm3 of solvent

  25. The solution of KOH has 70g in 500cm3 of water. Express the concentration in w/v% Answer: w/v means how many grams of KOH are in 100cm3 of water Given mass in 500cm3Want mass in 100cm3 70g of KOH in 500cm3 of water You need to divide by 5! 70/5 = 14 14g of KOH in 100cm3 of solution Answer = 14w/v%

  26. Q225 .A solution of sodium chloride consists of 75 grams in a 250cm3 volume of water. Express the concentration in w/v% Answer: w/v means how many grams of NaCl are in 100cm3 of water Given mass in 250cm3Want mass in 100cm3 75g of NaCl in 250cm3 75 / 250 x 100 = 30 30g of NaCl in 100cm3 of solution Answer = 30 w/v%

  27. 4. Volume per volume • Concentration is also sometimes measured in % volume of solute (cm3) per solvent (cm3) • This is written as % (v/v) • 5% (v/v) means 1cm3 of solute in 100cm3 of solvent

  28. USE: This is used for expressing the alcohol content in wine • 11% written on a bottle of wine means 11cm3 of alcohol in 100cm3 of wine!

  29. The label on a bottle of whiskey says that the ethanol content is 40% (v/v). How many cm3 of ethanol are there is 30cm3 of the whiskey? Answer: 40%v/v means 40cm3 of ethanol in 100cm3 of whiskey Given volume in 100cm3Want volume in 30cm3 • x cm3 of ethanol = 30cm3 ofwhiskey 40cm3 of ethanol = 100cm3 of whiskey (30)(40) = (100)(x) 1200 = x 100 12 = x 12cm3 of ethanol in 30cm3 of whiskey

  30. Q226.The label on a bottle of beer says that the ethanol content is 4.8% (v/v). How many cm3 of ethanol are there in one pint 568cm3 of the beer? Answer: 4.8%v/v means 4.8cm3 of ethanol in 100cm3 of beer Given volume in 100cm3Want volume in 568cm3 (568)(4.8) / (100) = 27.264 27.264 cm3 of ethanol in 30cm3 of beer

  31. 5. Mass per mass • Concentration is also sometimes measured in % mass of solute(g) per mass of solvent(g) • This is written as % (w/w) • 36% (w/w) means 36gof solute in 100gof solvent

  32. A solution has a w/w% concentration of 35%. What mass of solute would be present in 450g of solvent? Answer: 35% w/w means 35g of solute in 100g of solvent Given mass in 100gWant mass in 450g (35)/100 x 450 = ? (35)/100 x 450 = 157.5 157.5g of solute in 450g of solvent Answer: 157.5g

  33. Q227. Concentrated hydrochloric acid is a 37.9 (w/w)% solution of HCl in water. What mass of this solution would have to be taken so that it contains 5g of HCl? Answer: 37.9% w/w means 39.7g of solute in 100g of solvent Given mass in 100gWant mass of solvent with 5g of solute 37.9/ 100 = 0.379g in 1cm3 of solvent 5/0.379 = 13.1926 There would need to be 13.1926g of solution Answer: 13.1936g

  34. Standard solutions • A standard solution is a solution whose concentration is accurately known

  35. Primary Standard solution • A primary standard is a water soluble substance that is stable and available in pure form. • They are used to make standard solutions.

  36. Examples of Primary Standards Sodium Chloride Anhydrous Sodium Carbonate Potassium Dichromate

  37. Common substances that are not primary standards include: Sulfuric acid – absorbs water vapour from the air. Sodium hydroxide – absorbs water vapour from the air. Iodine – sublimes.

  38. Preparation of a standard solution of sodium carbonate. We want to make up 250cm3 of a 0.1M solution of Na2CO3 What mass of sodium carbonate is needed??

  39. How to make a standard solution • A special technique is used so that the concentration of the solution can be accurately known! • Find the mass of the solute accurately using a mass balance, and a clock glass.

  40. 2. The solute is transferred to a clean beaker (filled with some deionised water) using a stirring rod

  41. 3. A wash bottle (filled with deionised water) is used to rinse the clock glass and stirring rod into the beaker. The solution is stirred to dissolve the solute completely.

  42. 5. The solution from the beaker is poured into a volumetric flask

  43. 6. Any solution remaining in the beaker is washed into the volumetric flask, using a wash bottle

  44. 7. The flask is filled up to close to the calibration mark with a wash bottle, and then drop wise until the bottom of the meniscus is at the calibration mark.

  45. 8. The flask is stoppered, and inverted 20 times to sure the solution is mixed properly!

  46. 6. Expressing concentration in parts per million (ppm) • ppm (parts per million) = milligrams in a litre

  47. Convert the following to ppm0.62g / 500cm3 Given grams per 500cm3 Grams per 1000cm3 Need: mg per 1000cm3 0.62g in 500cm3 (0.62) / 500 x 1000 = 1.24g 1.24g in 1000cm3 To convert grams to milligrams 1.24 x 1000 = 1240 1240 mg in 1000cm3 Answer = 1240ppm

  48. 228(a)Convert the following to ppm0.54g / 1L Grams per 1000cm3 Need: mg per 1000cm3 To convert grams to milligrams multiply by 1000 0.54g x 1000 = 540 540 mg in 1000cm3 Answer = 540ppm

  49. 228(b)Convert the following to ppm0.18g / 1L Grams per 1000cm3 Need: mg per 1000cm3 To convert grams to milligrams 0.18g x 1000 = 180 180 mg in 1000cm3 Answer = 180ppm

  50. 228c)Convert the following to ppm0.077g / 100cm3 Given grams per 100cm3 Grams per 1000cm3 Need: mg per 1000cm3 (0.077) x 10 = 0.77 0.77g in 1000cm3 To convert grams to milligrams (1000)(0.77) = 770 770 mg in 1000cm3 Answer = 770ppm

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