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- 1. CONCENTRATION OFSOLUTIONS
- 2. Concentration = amount of solute perquantity of solventMass/volume % = Mass of solute (g) x 100%Volume of solution (mL)CONCENTRATION AS A MASS/VOLUME PERCENTUsually for solids dissolved in liquids
- 3. SAMPLE PROBLEM:2.00mL of distilled water is added to 4.00g of apowdered drug. The final volume is 3.00mL. What is theconcentration of the drug in g/100mL of solution? Whatis the percent (m/v) of the solution?CONCENTRATION AS A MASS/VOLUME PERCENT
- 4. = (4.00g) x 100%(3.00mL)CONCENTRATION AS A MASS/VOLUME PERCENTSAMPLE PROBLEM:2.00mL of distilled water is added to 4.00g of a powdered drug. The final volumeis 3.00mL. What is the concentration of the drug in g/100mL of solution? What isthe percent (m/v) of the solution?Mass/volume % = Mass of solute x 100%Volume of solutionTherefore the mass/volume percent is 133%= 133%
- 5. 1.7% = Mass of solute x 100%(2000mL)SAMPLE PROBLEM:Many people use a solution of Na3PO4 to clean walls beforeputting up wallpaper. The recommended concentration is 1.7%(m/v) . What mass ofNa3PO4 is needed to make 2.0L of solution?CONCENTRATION AS A MASS/VOLUME PERCENTMass/volume % = Mass of solute x 100%Volume of solutionMass of solute = 34gTherefore the mass required is 34g
- 6. CONCENTRATION AS A MASS/MASS PERCENTConcentration = amount of solute perquantity of solventMass/Mass % = Mass of solute (g) x 100%Mass of solution (g)Usually for solids dissolved in liquids
- 7. Mass/Mass % = Mass of solute x 100%Mass of solutionSAMPLE PROBLEM:Calcium chloride, CaCl2, can be used instead of road salt to melt theice on roads during the winter. To determine how much calciumchloride had been used on a nearby road, a student took a sample ofslush to analyze. The sample had a mass of 23.47g. When thesolution was evapourated, the residue had a mass of 4.58g (assumethat no other solutes were present).a) What was the mass/mass percent of calcium chloride in the slush?b) How many grams of calcium chloride were present in 100g solution?CONCENTRATION AS A MASS/MASS PERCENT
- 8. Mass/Mass % = Mass of solute x 100%Mass of solution= (4.58g) x 100%(23.47g)= 19.5%Therefore the mass/mass percent of calcium chloride is19.5%SAMPLE PROBLEM:Calcium chloride, CaCl2, can be used instead of road salt to melt the ice on roads during the winter. Todetermine how much calcium chloride had been used on a nearby road, a student took a sample ofslush to analyze. The sample had a mass of 23.47g. When the solution was evapourated, the residue hada mass of 4.58g (assume that no other solutes were present).a) What was the mass/mass percent of calcium chloride in the slush?CONCENTRATION AS A MASS/MASS PERCENT
- 9. If your mass/mass % of CaCl2 is 19.5% and youhave 100g, then you have 19.5g of CaCl2Ex: 19.5% x 100g = 19.5gSAMPLE PROBLEM:Calcium chloride, CaCl2, can be used instead of road salt to melt the ice on roads during the winter. Todetermine how much calcium chloride had been used on a nearby road, a student took a sample ofslush to analyze. The sample had a mass of 23.47g. When the solution was evapourated, the residue hada mass of 4.58g (assume that no other solutes were present).b) How many grams of calcium chloride were present in 100g solution?CONCENTRATION AS A MASS/MASS PERCENT
- 10. Concentration = amount of solute perquantity of solventVolume/Volume % = Volume of solute (mL) x 100%Volume of solution (mL)CONCENTRATION AS A VOLUME/VOLUME PERCENTUsually for liquids dissolved in liquids
- 11. CONCENTRATION AS A VOLUME/VOLUME PERCENTSAMPLE PROBLEM:Rubbing alcohol is sold as a 70% (v/v) solution of isopropyl alcohol inwater. What volume of isopropyl alcohol is used to make 500mL ofrubbing alcohol?Volume/Volume % = volume of solute x 100%volume of solution(70%) = volume of solute x 100%(500mL)= 350mLTherefore the volume is 350mL
- 12. CONCENTRATION AS PARTS PER MILLION (ppm)ppm = Mass of solute (g) x 106Mass of solution (g)Usually mass/mass relationships BUT does not refer tothe number of particlesMass of solute (g) = xMass of solution (g) 106g of solutionOR
- 13. CONCENTRATION AS PARTS PER BILLION (ppB)ppb = Mass of solute (g) x 109Mass of solution (g)To the power of 9 instead…Mass of solute (g) = xMass of solution (g) 109g of solutionOR
- 14. CONCENTRATION AS PARTS PER BILLION (ppB)SAMPLE PROBLEM:A fungus that grows on peanuts produces a deadly toxin.When ingested in large amounts, this toxin destroys the liverand can cause cancer. Any shipment of peanuts that containsmore than 25ppb of this fungus is rejected. A companyreceives 20 tonnes of peanuts to make peanut butter. What isthe maximum mass (in g) of fungus that is allowed?
- 15. CONCENTRATION AS PARTS PER BILLION (ppB)SAMPLE PROBLEM:A fungus that grows on peanuts produces a deadly toxin. When ingested in large amounts,this toxin destroys the liver and can cause cancer. Any shipment of peanuts that containsmore than 25ppb of this fungus is rejected. A company receives 20 t of peanuts to makepeanut butter. What is the maximum mass (in g) of fungus that is allowed?ppb = Mass of fungus (g) x 109Mass of peanuts (g)Convert 20 t to grams:20t x 1000kg/t x 1000g/kg = 20 x 106g25ppb = Mass of fungus x 10920 x 106g25ppb = Mass of fungus x 10320500 = Mass of fungus1030.5g = Therefore the maximum mass of thefungus allowed is 0.5g
- 16. MOLAR CONCENTRATION
- 17. MOLAR CONCENTRATIONThe number of moles of solute in 1L of solution(this is the standard measure of concentration in chemistry!!!)Molar concentration (mol/L) = moles of soluteVolume of solution (L)C = nVSimplified…
- 18. MOLAR CONCENTRATIONSAMPLE PROBLEM:A saline solution contains 0.90g of NaCl dissolved in 100mLof solution. What is the molar concentration of the solution?C = nVGiven:m = 0.90gn = m/M = 0.90g / 58.44g/mol = 0.0154molV = 100mL = 0.1LC = (0.0154mol)(0.1L)= 0.154mol/LTherefore the molar concentration of the saline solution is 0.15 mol/L
- 19. MOLAR CONCENTRATIONSAMPLE PROBLEM:At 20°C, a saturated solution of CaSO4 has a concentration of0.0153mol/L. A student takes 65mL of this solution andevaporates it. What mass (in g) is left in the evapouratingdish?C = nVGiven:C = 0.0153mol/LV = 65mL = 0.065Ln = ? 0.0153mol/L = n(0.065L)= 0.000994molNOT DONE YET! Convert to grams!MCaSO4= 136.15g/molmCaSO4= n x M= 0.000994mol x 136.15g/mol= 0.135gTherefore 0.14g ofCaSO4 are left in theevapourating dish

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