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 & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter13: Solutions And Their Behavior

Section: Chapter Questions

Problem 106SCQ

See similar textbooks

Similar questions

A gaseous solute dissolves in water. The solution process has H=15 kJ. Its solubility at 22C and 6.00 atm is 0.0300 M. Would you expect the solubility to be greater or less at (a) 22C and 1 atm? (a) 18C and 6 atm? (a) 15C and 10 atm? (a) 35C and 3 atm?
The Henry's law constant for the solubility of radon in water at is 9.57106 M/mm Hg. Radon is present with other gases in a sample taken from an aquifer at 30C. Radon has a mole fraction of 2.7106 in the gaseous mixture. The gaseous mixture is shaken with water at a total pressure of 28 atm. Calculate the concentration of radon in the water. Express your answers using the following concentration units. (a) molarity (b) ppm (Assume that the water sample has a density of 1.00 g/mL.)
A 2.77 M NaOH solution in water has a density of 1.109 g/mL. Express the concentration of this solution (a) mass percentage. (b) mole fraction. (c) molality.
Calculate the freezing point and normal boiling points of each of the following aqueous solutions. (a) 2.63 m acetic acid (b) 33.0 % by mass lactose, C12H22O11 (c) 32.15 mL of ethylene glycol, C2H6O2(d=1.113g/mL) in 624 mL of water (d=1.00g/mL)
You have read that adding a solute to a solvent can both increase the boiling point and decrease the freezing point. A friend of yours explains it to you like this: The solute and solvent can be like salt in water. The salt gets in the way of freezing in that it blocks the water molecules from joining together. The salt acts like a strong bond holding the water molecules together so that it is harder to boil. What do you say to your friend?