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Concentration of solutions Parts per million (ppm) This is a way of expressing very dilute concentrations of substances. Just as per cent means out of.

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Presentation on theme: "Concentration of solutions Parts per million (ppm) This is a way of expressing very dilute concentrations of substances. Just as per cent means out of."— Presentation transcript:

1 Concentration of solutions Parts per million (ppm) This is a way of expressing very dilute concentrations of substances. Just as per cent means out of a hundred, so parts per million or ppm means out of a million. e.g. One ppm is equivalent to 1 milligram of something per liter of water (mg/l) or 1 milligram of something per kilogram soild (mg/kg). One part per million (ppm) denotes one part per 1,000,000 parts, one part in 10 6,million

2 UnitDefinitionApplication *The molarity of a solution is temperature dependent, but the other units shown in this table are independent of temperature. molarity (M)moles of solute/liter of solution (mol/L) Used for quantitative reactions in solution and titrations; mass and molecular mass of solute and volume of solution are known. mole fraction (X)moles of solute/total moles present (mol/mol) Used for partial pressures of gases and vapor pressures of some solutions; mass and molecular mass of each component are known. molality (m)moles of solute/kg of solvent (mol/kg) Used in determining how colligative properties vary with solute concentration; masses and molecular mass of solute are known. mass percentage (%)[mass of solute (g)/mass of solution (g)] × 100 Useful when masses are known but molecular masses are unknown. parts per thousand (ppt)[mass of solute/mass of solution] × 103 (g solute/kg solution) Used in the health sciences, ratio solutions are typically expressed as a proportion, such as 1:1000. parts per million (ppm)[mass of solute/mass of solution] × 106 (mg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown. parts per billion (ppb)[mass of solute/mass of solution] × 109 (µg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.

3 Molarity Molarity is the number of moles of solute in one liter of solution. Example: 2500 ml of solution that contains 90.0 g of sodium chloride NaCl. What is the Molarity of that solution? (molar mass Cl = 35.5 g/mol, Na = 23 g/mol )

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6 SOLUTION Molarity of solution = n moles of solutevolume of solution(l) n moles of solute = Molarity of solution × volume of solution(l) = 2.0 M × 1.5 L= 3.0 mol Molar Mass of solute = mass (g)/n moles of solute (mol) = 20 (g)/3.o(mol) = 6.66 g/mol Practice:

7 * Molality: Molality (symbolized by m): the number of moles of solute per kilogram of solvent. Example: A solution was prepared by dissolving 17.1 g of sucrose (table sugar) (C 12 H 22 O 11 ) in 125 g of water. Calculate the molal concentration of this solution. (Molar Mass of C 12 H 22 O 11 = 342 g/mol)

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9 SOLUTION n moles of iodine = Molality of solution × mass of solvent = 0.480 m × 0.1 Kg = 0.048 mol mass of iodine = n moles of iodine × Molar Mass of I 2 = 0.048 × 253.8 = 12.18g

10 the equivalent mass of sulfuric acid is the molar mass divided by 2, since each mole of H 2 SO 4 can furnish 2 moles of protons. The equivalent mass of calcium hydroxide is also half the molar mass, since each mole of Ca(OH) 2 contains 2 moles of OH - ions that can react with 2 moles of protons. The equivalent is defined so that 1 equivalent of acid will react with exactly 1 equivalent of base.

11 Example: The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution. Calculate the normality of the sulfuric acid. S

12 Example: A solution is prepared by mixing 5.00 g acetic acid (CH 3 COOH) with 100.0 g water. Calculate the mole fraction of acetic acid in this solution. (Molar Mass of CH 3 COOH = 60 g/mol, Molar Mass of water = 18.0 g/mol)

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14 SOLUTION Example: What is mass of solvent is needed to add to 7.50 g of solute to prepare solution its concentration 28.0 % by mass?

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18 Vodka is essentially a solution of pure ethanol in water. Typical vodka is sold as “80 proof,” which means that it contains 40.0% ethanol by volume. The density of pure ethanol is 0.789 g/mL at 20°C. If we assume that the volume of the solution is the sum of the volumes of the components (which is not strictly correct), calculate the following for the ethanol in 80-proof vodka. 1- the mass percentage 2- the mole fraction 3- the molarity 4- the molality


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