A. Freezing Point of Cyclohexane (Solvent) 1. Mass of beaker, test tube (g) 85.364g 2. Freezing point, from cooling curve (°C) 5.68°C B. Freezing Point of Cyclohexane plus Unknown Solute 1. Mass of beaker, test tube, cyclohexane (g) 97.594g 2.* Mass of cyclohexane (g) 3. Tared mass of added solute (g) 0.311g 4. Freezing point, from cooling curve (°C) 1.72°C Calculations 1. ke for cyclohexane (°Ckg/mol) 20.0°C-kg/mol 2. Freezing-point change AT:(°C) 3. Mass of cyclohexane in solution (kg) 4. Moles of solute, total (mol) 5. Mass of solute, total (g) 6. Molar mass of solute (g/mol) Based on your data (the molar of your unknown) what is the likely identity of the solute from the following selection: Naphthalene [C10Hg=128.171g/mol] Aspirin [C,H3O=180.158g/mol] Sucrose [C12H2201=342.3g/mol] Pentanoic acid [C;H19O2=102.13g/mol] 1-Pentanol [C§H120=88.15g/mol] Lactic acid [C3H03=90.08g/mol] 7. Identity of your unknown. Key Equations: ATu: change in the freezing point (°C) n: moles of solute m: molality Uolven' mass of solvent (kg) Molmai Molar of mass of solute ke freezing point depression of solvent i: Van't Hoff factor Uowwi mass of solute (g) ATK = i k,m m=n/mobouh AT = i kelowa/ Mi ATK =| Toow - Tuvw

A. Freezing Point of Cyclohexane (Solvent) 1. Mass of beaker, test tube (g) 85.364g 2. Freezing point, from cooling curve (°C) 5.68°C B. Freezing Point of Cyclohexane plus Unknown Solute 1. Mass of beaker, test tube, cyclohexane (g) 97.594g 2.* Mass of cyclohexane (g) 3. Tared mass of added solute (g) 0.311g 4. Freezing point, from cooling curve (°C) 1.72°C Calculations 1. ke for cyclohexane (°Ckg/mol) 20.0°C-kg/mol 2. Freezing-point change AT:(°C) 3. Mass of cyclohexane in solution (kg) 4. Moles of solute, total (mol) 5. Mass of solute, total (g) 6. Molar mass of solute (g/mol) Based on your data (the molar of your unknown) what is the likely identity of the solute from the following selection: Naphthalene [C10Hg=128.171g/mol] Aspirin [C,H3O=180.158g/mol] Sucrose [C12H2201=342.3g/mol] Pentanoic acid [C;H19O2=102.13g/mol] 1-Pentanol [C§H120=88.15g/mol] Lactic acid [C3H03=90.08g/mol] 7. Identity of your unknown. Key Equations: ATu: change in the freezing point (°C) n: moles of solute m: molality Uolven' mass of solvent (kg) Molmai Molar of mass of solute ke freezing point depression of solvent i: Van't Hoff factor Uowwi mass of solute (g) ATK = i k,m m=n/mobouh AT = i kelowa/ Mi ATK =| Toow - Tuvw

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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The freezing point of water is 0.00 °C at 1 atmosphere. A student dissolves 11.60 grams of cobalt(II) sulfate CoSO4, (155.0 g/mol), in 279.3 grams of water. Use the table of boiling and freezing point constants to answer the questions below. Solvent Formula Kb(C/m) Kf(C/m) Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67 Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02 The molality of the solution is m. The freezing point of the solution is °C.
The freezing point of water is 0.00 °C at 1 atmosphere. A student dissolves 11.60 grams of cobalt(II) sulfate CoSO4, (155.0 g/mol), in 279.3 grams of water. Use the table of boiling and freezing point constants to answer the questions below. Solvent Formula Kb(C/m) Kf(C/m) Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67 Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02 The molality of the solution is m. The freezing point of the solution is °C.
Calculate the molalities of the following aqueous solutions. Part 1 of 3 1.13 M sugar (C₁2H₂2011) solution (density of solution = 1.12 II Part 2 of 3 m m Part 3 of 3 0.4.2 X 0.791 M NaOH solution (density of solution = 1.04). Round your answer to 3 significant digits. mL m S -). Round your answer to 2 significant digits. mL 5.37 M NaHCO3 solution (density of solution = 1.19). Round your answer to 2 significant digits. mL X
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The freezing point of benzene, C6H₁, is 5.50 °C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT. A student dissolves 11.78 grams of DDT, C14H9 C15 (354.5 g/mol), in 252.1 grams of benzene. Use the table of boiling and freezing point constants to answer the questions below. Solvent Water Ethanol Chloroform Benzene Formula H₂O CH3 CH₂ OH CHC13 C6H6 Diethyl ether CH3 CH₂ OCH2 CH3 The molality of the solution is The freezing point of the solution is Kb (°C/m) Kf (°C/m) 1.86 1.99 0.512 1.22 3.67 2.53 2.02 m. °C. 5.12
Calculate the freezing point of a solution containing 74.0 g of eucalyptol (C10H8O) solute dissolved in 0.600 kg of CHCl3. The freezing point of pure CHCl3 is -63.5 oC. CHCl3 has a Kf of 4.68 oC/molality. -64.9 oC -67.5 oC -48.3 oC -58.0 oC -69.0 oC -62.1 oC -59.5 oC -78.7 oC
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The freezing point of pure benzene is 5.5°C. The freezing point of a solution in which benzene is the solvent could be: a) 12.5°C b) 4.8°C Oc) 6.2°C d) 5.5°C