Properties of Solutions

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Presentation transcript:

Properties of Solutions Lesson 2

Measuring concentration Concentration = amount of solute/amount of solution Weight/volume percent Mass solute in g/volume of soln in mL x 100% Weight/weight percent Mass solute in g/mass solution in g x 100%

Moles of solute/Liters of Solution (M) Molality = Concentration Molarity = Moles of solute/Liters of Solution (M) Molality = Moles of solute/Kg of Solvent (m) Mole Fraction= Moles solute/total number of moles Mass %= Mass solute/total mass x 100 Copyright © Houghton Mifflin Company. All rights reserved.

Moles of solute/Liters of Solution (M) Molality = Concentration Molarity = Moles of solute/Liters of Solution (M) Molality = Moles of solute/Kg of Solvent (m) Mole Fraction= Moles solute/total number of moles Mass %= Mass solute/total mass x 100 Copyright © Houghton Mifflin Company. All rights reserved.

Moles of solute/Liters of Solution (M) Molality = Concentration Molarity = Moles of solute/Liters of Solution (M) Molality = Moles of solute/Kg of Solvent (m) Mole Fraction= Moles solute/total number of moles Mass %= Mass solute/total mass x 100 Copyright © Houghton Mifflin Company. All rights reserved.

Molarity

Convert the given grams of solute to moles of solute : A sample of NaNO3 weighing 8.5 grams is placed in a 500 ml volumetric flask and distilled water was added to the mark on the neck of the flask. Calculate the Molarity of the resulting solution. Convert the given grams of solute to moles of solute : Convert given ml of solution to liters Apply the definition for Molarity: Molarity = moles NaNO3 / volume of the solution in liters M = 0.1 mole / .500 liters = 0.200 Molar NaNO3 Copyright © Houghton Mifflin Company. All rights reserved.

Exercise #1 You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity. 8.00 M

Exercise #2 You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar? 0.200 L

Exercise #3 Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the concentration of each solution in units of molarity. 10.0 M NaOH 5.37 M KCl

Moles of solute/Liters of Solution (M) Molality = Molarity = Moles of solute/Liters of Solution (M) Molality = Moles of solute/Kg of Solvent (m) Mole Fraction= Moles solute/total number of moles Mass %= Mass solute/total mass x 100 Copyright © Houghton Mifflin Company. All rights reserved.

Molality

Exercise #4 A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.) 0.816 m

Moles of solute/Liters of Solution (M) Molality = Molarity = Moles of solute/Liters of Solution (M) Molality = Moles of solute/Kg of Solvent (m) Mole Fraction= Moles solute/total number of moles Mass %= Mass solute/total mass x 100 Copyright © Houghton Mifflin Company. All rights reserved.

Determine the mole fraction of KCl in 3000 grams of aqueous solution containing 37.3 grams of Potassium Chloride KCl. Convert grams KCl to moles KCl using the molecular weight of KCl 2. Determine the grams of pure solvent water from the given grams of solution and solute Total grams = 3000 grams = Mass of solute + Mass of water Mass of pure solvent = (3000 - 37.3) gram = 2962.7 gram Copyright © Houghton Mifflin Company. All rights reserved.

Determine the mole fraction of KCl in 3000 grams of aqueous solution containing 37.3 grams of Potassium Chloride KCl. 3. Convert grams of solvent H2O to mols 4. Apply the definition for mole fraction mole fraction = moles of KCl / Total mols of KCl and water = 0.5 / (0.5 + 164.6) = 0.5 / 165.1 = 0.00303 Copyright © Houghton Mifflin Company. All rights reserved.

Mole Fraction

Exercise #5 A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the mole fraction of H3PO4. (Assume water has a density of 1.00 g/mL.) 0.0145

Mass Percent

Exercise #6 What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water? 6.6%

Assuming the density of water to be 1 g/mL we approximate the density of a dilute aqueous solution to be 1 g/mL  1 ppm = 1 μg/mL = 1 mg/L  1 ppb = 1 ng/mL = 1 μg/L Copyright © Houghton Mifflin Company. All rights reserved.

Molarity Concentration is usually expressed in terms of molarity: Moles of solute/liters of solution (M) Moles of solute = molarity x volume of solution Moles = M x V

Molarity and concentration M = moles solute/liter of solution Dilution M1V1 = M2V2

Example What is molarity of 50 ml solution containing 2.355 g H2SO4? Molar mass H2SO4 = 98.1 g/mol Moles H2SO4 = .0240 mol (2.355 g/98.1 g/mol) Volume of solution = 50 mL/1000 mL/L = .050 L Concentration = moles/volume = .0240 mol/.050 L = 0.480 M

Solution stoichiometry How much volume of one solution to react with another solution Given volume of A with molarity MA Determine moles A Determine moles B Find target volume of B with molarity MB