Introductory Chemistry For Today
Introductory Chemistry For Today
8th Edition
ISBN: 9781285644561
Author: Seager
Publisher: Cengage
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Solutions
Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to a…
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Apply algebraic properties, formulas, and relationships to perform operations on real-world problems (e.g., solve for: molality. • Molality is the number of moles of solute dissolved in a kilogram of solvent. molality = moles of solute or molal, m kg of solvent Sample Problem #1: What is the molality of the example sucrose solution? First Step: Known and Unknown Quantities known unknown mass of sucrose = 0.030464 g C12H2201 molality C12H2201 = ? m mass of water = 0.981256 g H,0 Second Step: Formula molality of C12H22O11 = mol of C12H22O1 1 kg of H,0 Third Step: Rearranging, intermediate calculations, and unit conversions No rearranging required. Intermediate calculations: molar mass of C12H2201 = (12 x 12.011 g) + (22 x 1.0079 g) + (11 x 15.9994 g) = 342.299 g # mol C12H2201 = 0.030464 g C12H,O1 x 1 mol C,,H„01 = 0.000089 mol C12H22O1 1 342.299 g C12H2201 Unit conversions: 0.981256 g H,Ox_1.00000 kg_ = 0.000981256 kg H,0 1 1000.00 g Fourth Step: DELL 「田
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Transcribed Image Text:Apply algebraic properties, formulas, and relationships to perform operations on real-world problems (e.g., solve for: molality. • Molality is the number of moles of solute dissolved in a kilogram of solvent. molality = moles of solute or molal, m kg of solvent Sample Problem #1: What is the molality of the example sucrose solution? First Step: Known and Unknown Quantities known unknown mass of sucrose = 0.030464 g C12H2201 molality C12H2201 = ? m mass of water = 0.981256 g H,0 Second Step: Formula molality of C12H22O11 = mol of C12H22O1 1 kg of H,0 Third Step: Rearranging, intermediate calculations, and unit conversions No rearranging required. Intermediate calculations: molar mass of C12H2201 = (12 x 12.011 g) + (22 x 1.0079 g) + (11 x 15.9994 g) = 342.299 g # mol C12H2201 = 0.030464 g C12H,O1 x 1 mol C,,H„01 = 0.000089 mol C12H22O1 1 342.299 g C12H2201 Unit conversions: 0.981256 g H,Ox_1.00000 kg_ = 0.000981256 kg H,0 1 1000.00 g Fourth Step: DELL 「田
Fourth Step: molality, C12H2201 = 0.000089 mol C1,HO1 = 0.091 m 0.000981256 kg H20 Sample Problem #2: What is the molality of a solution of 0.20 g of benzoic acid, C,H,O2, in 20.0 g of naphthalene? moles of C,HO2 = 0.20 g 122 g/mol = 0.0016 mol C,H¢O2 molality of C,HO2 = 0.0016 mol C,H;O> 0.020 kg of naphthalene = 0.080 molal (mol/kg) Note: The "molal" unit is mol of solute/kg of solvent. The molal unit is abbreviated with the lower case letter, m. The molality of the C;HgO2 solution is 0.080 m. Practice Problem #1: What is the molality of a solution of 0.50 g of glucose, CH12O6, 14.3 g of water? O 35 m O 0.35 m O 0.19 m O 0.00019 m DELL & 司 %24 #3
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Transcribed Image Text:Fourth Step: molality, C12H2201 = 0.000089 mol C1,HO1 = 0.091 m 0.000981256 kg H20 Sample Problem #2: What is the molality of a solution of 0.20 g of benzoic acid, C,H,O2, in 20.0 g of naphthalene? moles of C,HO2 = 0.20 g 122 g/mol = 0.0016 mol C,H¢O2 molality of C,HO2 = 0.0016 mol C,H;O> 0.020 kg of naphthalene = 0.080 molal (mol/kg) Note: The "molal" unit is mol of solute/kg of solvent. The molal unit is abbreviated with the lower case letter, m. The molality of the C;HgO2 solution is 0.080 m. Practice Problem #1: What is the molality of a solution of 0.50 g of glucose, CH12O6, 14.3 g of water? O 35 m O 0.35 m O 0.19 m O 0.00019 m DELL & 司 %24 #3
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