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Two aqueous solutions are prepared in which one contains 0.275 moles of the nonvolatile solut glucose (C6H12O6) dissolved in 2500.00 g of water and the other solution contains 0.275 moles of the nonvolatile solute CaCl2 dissolved in 2500.00 g of water. Describe how each of the colligative properties listed below would be affected for each of these solutions (for each property, describe if there is a decrease or an increase then identify which of the solutions (C6H12O6 or CaCl2) would cause the greater effect). In addition, choose only 1 of the colligative properties (choose either the melting point or boiling point) and show how to determine (and then calculate) the magnitude of the change and resulting value for either the melting point or boiling point for each of these solutions (follow worked example 9.15 and 9.16 in 4. your text). Show how to determine (and then calculate) the concentration of each solution in both % (w/w) and molaliy. The molality of each solution is: The % (w/w) of the glucose (CoH12O6) solution is: The % (w/w) of the calcium chloride (CaCl2) solution is: melting point of each solution (the normal melting point of H2O is 0.000 C & Kf 1.8600 °C kg /mol) (CoH12O6 or CaCl)_ solution. solution would be (#)times_(lower or higher) The melting point of the than the (C6H12O6 or CaCl) ATr iKfem and the new freezing point would be: ATf for CoH12O6 is: iKfm and the new freezing point would be: ATf ATr for CaCl2 is: boiling point of each solution (the normal boiling point of H20 is 100.000 C & Kb = 0.5100 C kg / mol)) solution would be_(#)_times_(lower or higher) The boiling point of the than the_(CoH12O6 or CaCl)_solution. (C6H1206 or CaCl) i-Kf 'm and the new boiling point would be: ATf ATr for CoH12O6 is: i-Krm and the new boiling point would be: ATf AT for CaCl2 is: атe 5. An aqueous solution is prepared by dissolving 0 solution is then diluted by taking a 25.0 mL aliquot A third solution is then prepared by taking 10.0 mL or 50.0 mL with water. Show how to determine (and then 3 different solutions prepared)

Two aqueous solutions are prepared in which one contains 0.275 moles of the nonvolatile solut glucose (C6H12O6) dissolved in 2500.00 g of water and the other solution contains 0.275 moles of the nonvolatile solute CaCl2 dissolved in 2500.00 g of water. Describe how each of the colligative properties listed below would be affected for each of these solutions (for each property, describe if there is a decrease or an increase then identify which of the solutions (C6H12O6 or CaCl2) would cause the greater effect). In addition, choose only 1 of the colligative properties (choose either the melting point or boiling point) and show how to determine (and then calculate) the magnitude of the change and resulting value for either the melting point or boiling point for each of these solutions (follow worked example 9.15 and 9.16 in 4. your text). Show how to determine (and then calculate) the concentration of each solution in both % (w/w) and molaliy. The molality of each solution is: The % (w/w) of the glucose (CoH12O6) solution is: The % (w/w) of the calcium chloride (CaCl2) solution is: melting point of each solution (the normal melting point of H2O is 0.000 C & Kf 1.8600 °C kg /mol) (CoH12O6 or CaCl)_ solution. solution would be (#)times_(lower or higher) The melting point of the than the (C6H12O6 or CaCl) ATr iKfem and the new freezing point would be: ATf for CoH12O6 is: iKfm and the new freezing point would be: ATf ATr for CaCl2 is: boiling point of each solution (the normal boiling point of H20 is 100.000 C & Kb = 0.5100 C kg / mol)) solution would be_(#)_times_(lower or higher) The boiling point of the than the_(CoH12O6 or CaCl)_solution. (C6H1206 or CaCl) i-Kf 'm and the new boiling point would be: ATf ATr for CoH12O6 is: i-Krm and the new boiling point would be: ATf AT for CaCl2 is: атe 5. An aqueous solution is prepared by dissolving 0 solution is then diluted by taking a 25.0 mL aliquot A third solution is then prepared by taking 10.0 mL or 50.0 mL with water. Show how to determine (and then 3 different solutions prepared)

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Two aqueous solutions are prepared in which one contains 0.275 moles of the nonvolatile solut glucose (C6H12O6) dissolved in 2500.00 g of water and the other solution contains 0.275 moles of the nonvolatile solute CaCl2 dissolved in 2500.00 g of water. Describe how each of the colligative properties listed below would be affected for each of these solutions (for each property, describe if there is a decrease or an increase then identify which of the solutions (C6H12O6 or CaCl2) would cause the greater effect). In addition, choose only 1 of the colligative properties (choose either the melting point or boiling point) and show how to determine (and then calculate) the magnitude of the change and resulting value for either the melting point or boiling point for each of these solutions (follow worked example 9.15 and 9.16 in 4. your text). Show how to determine (and then calculate) the concentration of each solution in both % (w/w) and molaliy. The molality of each solution is: The % (w/w) of the glucose (CoH12O6) solution is: The % (w/w) of the calcium chloride (CaCl2) solution is: melting point of each solution (the normal melting point of H2O is 0.000 C & Kf 1.8600 °C kg /mol) (CoH12O6 or CaCl)_ solution. solution would be (#)times_(lower or higher) The melting point of the than the (C6H12O6 or CaCl) ATr iKfem and the new freezing point would be: ATf for CoH12O6 is: iKfm and the new freezing point would be: ATf ATr for CaCl2 is: boiling point of each solution (the normal boiling point of H20 is 100.000 C & Kb = 0.5100 C kg / mol)) solution would be_(#)_times_(lower or higher) The boiling point of the than the_(CoH12O6 or CaCl)_solution. (C6H1206 or CaCl) i-Kf 'm and the new boiling point would be: ATf ATr for CoH12O6 is: i-Krm and the new boiling point would be: ATf AT for CaCl2 is: атe 5. An aqueous solution is prepared by dissolving 0 solution is then diluted by taking a 25.0 mL aliquot A third solution is then prepared by taking 10.0 mL or 50.0 mL with water. Show how to determine (and then 3 different solutions prepared)
Transcribed Image Text:Two aqueous solutions are prepared in which one contains 0.275 moles of the nonvolatile solut glucose (C6H12O6) dissolved in 2500.00 g of water and the other solution contains 0.275 moles of the nonvolatile solute CaCl2 dissolved in 2500.00 g of water. Describe how each of the colligative properties listed below would be affected for each of these solutions (for each property, describe if there is a decrease or an increase then identify which of the solutions (C6H12O6 or CaCl2) would cause the greater effect). In addition, choose only 1 of the colligative properties (choose either the melting point or boiling point) and show how to determine (and then calculate) the magnitude of the change and resulting value for either the melting point or boiling point for each of these solutions (follow worked example 9.15 and 9.16 in 4. your text). Show how to determine (and then calculate) the concentration of each solution in both % (w/w) and molaliy. The molality of each solution is: The % (w/w) of the glucose (CoH12O6) solution is: The % (w/w) of the calcium chloride (CaCl2) solution is: melting point of each solution (the normal melting point of H2O is 0.000 C & Kf 1.8600 °C kg /mol) (CoH12O6 or CaCl)_ solution. solution would be (#)times_(lower or higher) The melting point of the than the (C6H12O6 or CaCl) ATr iKfem and the new freezing point would be: ATf for CoH12O6 is: iKfm and the new freezing point would be: ATf ATr for CaCl2 is: boiling point of each solution (the normal boiling point of H20 is 100.000 C & Kb = 0.5100 C kg / mol)) solution would be_(#)_times_(lower or higher) The boiling point of the than the_(CoH12O6 or CaCl)_solution. (C6H1206 or CaCl) i-Kf 'm and the new boiling point would be: ATf ATr for CoH12O6 is: i-Krm and the new boiling point would be: ATf AT for CaCl2 is: атe 5. An aqueous solution is prepared by dissolving 0 solution is then diluted by taking a 25.0 mL aliquot A third solution is then prepared by taking 10.0 mL or 50.0 mL with water. Show how to determine (and then 3 different solutions prepared)
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