The freezing point of 53.37 g of a pure solvent is measured to be 49.20 ºC....
The freezing point of 53.37 g of a pure solvent is measured to be 49.20 ºC. When 2.74 g of an unknown solute (assume the van 't Hoff factor = 1.0000) is added to the solvent the freezing point is measured to be 47.05 ºC. Answer the following questions ( the freezing point depression constant of the pure solvent is 7.83 ºC·kg solvent/mol solute).
What is the molality of the solution? m
How many moles of solute are present? mol
What is the molecular weight of the solute? g/mol
The freezing point of 41.63 g of a pure solvent is measured to be 46.04 ºC. When 2.39 g of an unknown solute (assume the van 't Hoff factor = 1.0000) is added to the solvent the freezing point is measured to be 43.95 ºC.
If the freezing point depression constant of the pure solvent is 7.66 ºC·kg solvent/mol solute calculate the molar mass of the solute. g/mol
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The freezing point of 53.37 g of a pure solvent is measured to
be 49.20 ºC....
The freezing point of 53.48 g of a pure solvent is measured to be 41.91 ºC....
The freezing point of 53.48 g of a pure solvent is measured to be 41.91 ºC. When 2.94 g of an unknown solute (assume the van 't Hoff factor = 1.0000) is added to the solvent the freezing point is measured to be 39.86 ºC. Answer the following questions ( the freezing point depression constant of the pure solvent is 7.33 ºC·kg solvent/mol solute). -What is the molality of the solution?___ m -How many moles of solute are present? ___mol...
A freezing point depression experiment was conducted using tert-butyl alcohol as the solvent instead of water...
A freezing point depression experiment was conducted using tert-butyl alcohol as the solvent instead of water The freezing point of pure tert-butyl alcohol is 27C and Kf=9.1C/mol. To 25g of tert-butyl alcohol was added 0.70 g of a non-electrolyte solute (i.e. sugar). The solution was cooled and the new freezing point was determined to be 20C. 1. What is the molar mass of the unknown non-electrolyte solute? 2. In a similar experiment, MgCl2 (MW= 95.2 g/mol) solute was added to...
Solute X (molecular substance) was added to a pure solvent and the freezing point depression was...
Solute X (molecular substance) was added to a pure solvent and the freezing point depression was measured to be 7.0°C. Separately, the same amount of solute Y (molecular substance) was added to the same amount of pure solvent and the freezing point depression was measured to be 10.5°C. Which solute (X or Y) has the higher molar mass. Explain your rationale. thank you
Post-Lab Material Experiment 19 Data and Calculations: Molar Mass Determination by Depression of the Freezing Point...
Post-Lab Material Experiment 19 Data and Calculations: Molar Mass Determination by Depression of the Freezing Point 0.3 Name Section A. Measured Freezing Point of Pure Water B. Finding the Freezing Point of a Solution of Liquid Unknown Target mass of solute (Calculated based on the parameters in the instructions) Unknown # Liquid 3.5 Actual mass of solute used Trial Freezing point of solution (observed) -3.0 Mass of solution 116.6 Trial II Freezing point of solution -3.3 Mass of solution 101.3....
1h. A certain pure solvent freezes at 39.8°C and has a freezing point depression constant Kf...
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A pure solvent freezes at 6.35 degrees Celsius and a pure solute freezes at 832.0 degrees...
A pure solvent freezes at 6.35 degrees Celsius and a pure solute freezes at 832.0 degrees Celsius. The molar mass of the solvent is 96.4 g/mol and the molar mass of the solute is 78.5 g/mol. The freezing point depression constant for the solvent is 6.33 degrees/m. A solution is made of 1.012 g of the solute and 15.280 g of the solvent, determine its freezing point in degrees Celsius.
Drop at freezing point: A student performed the colligative properties experiment for which he measured 50.00mL...
Drop at freezing point: A student performed the colligative properties experiment for which he measured 50.00mL of a solvent and weighed it giving a mass of 75.25g. I determined that the freezing point of the pure solvent was 1.00C. Then he added 0.30g of an unknown solid and determined that the freezing point of solution was -1.00C. The Kf of the pure solvent is 20.2C / m. Use the information presented above to answer the following questions: 1. What is...
Freezing point depression can be used to experimentally determine the van't Hoff factor of a solute...
Freezing point depression can be used to experimentally determine the van't Hoff factor of a solute in solution. Given the data in the table, please answer the questions below and determine the "real" van't Hoff factor of the solute. Experimental Results Mass of solvent (water) Freezing point of water Freezing point depression constant (Kf) of water Mass of solution Freezing point of solution 8.515 g 0.00°C 1.86°C/m 9.3589 -5.45°C a. What mass of solute was used? b. What is the...
van't Hoff Factor Chem 202 Freezing Point of Aqueous Solutions Results Experimental freezing point values determined...
van't Hoff Factor
Chem 202 Freezing Point of Aqueous Solutions Results Experimental freezing point values determined from graphs (sce directions on p. 3-5, steps 2,3): Sample Ty measured /°C AT measured / °C Distilled Water (solvent) -0.1°C Follow your instructor's Solution D (Naci) -1.8°C 1.7°C directions for submitting these values Solution G (lalla) -1.0°C 0.9°C before you leave lab! van't Hoff factor values (see directions on p. 3-6, steps 4-5): Solution D Solute Compound Naci Molality 0.505 mol/kg Ideal van't...
Molar mass determination by depression of freezing point lab I'm stuck on calculating the moles of...
Molar mass determination by depression of freezing point lab I'm
stuck on calculating the moles of solute.. How do I calculate it?
Also can u please check if I've done everything else correctly..
The data I collected: Measured freezing point of pure water: 0.0
degrees Celsius Actual mass of solute used: 10.12g Freezing point
of solution (observed): -3.4 Celsius Mass of solution: 84.7g
Freezing point of a Solution of liqud unknown Freezing point depression: Trial #1. 0.0℃ (-3.4°C)= 3.4℃ Molality...
Post-Lab Material Experiment 19 Data and Calculations: Molar Mass Determination by Depression of the Freezing Point 0.3 Name Section A. Measured Freezing Point of Pure Water B. Finding the Freezing Point of a Solution of Liquid Unknown Target mass of solute (Calculated based on the parameters in the instructions) Unknown # Liquid 3.5 Actual mass of solute used Trial Freezing point of solution (observed) -3.0 Mass of solution 116.6 Trial II Freezing point of solution -3.3 Mass of solution 101.3....
Freezing point depression can be used to experimentally determine the van't Hoff factor of a solute in solution. Given the data in the table, please answer the questions below and determine the "real" van't Hoff factor of the solute. Experimental Results Mass of solvent (water) Freezing point of water Freezing point depression constant (Kf) of water Mass of solution Freezing point of solution 8.515 g 0.00°C 1.86°C/m 9.3589 -5.45°C a. What mass of solute was used? b. What is the...
van't Hoff Factor
Chem 202 Freezing Point of Aqueous Solutions Results Experimental freezing point values determined from graphs (sce directions on p. 3-5, steps 2,3): Sample Ty measured /°C AT measured / °C Distilled Water (solvent) -0.1°C Follow your instructor's Solution D (Naci) -1.8°C 1.7°C directions for submitting these values Solution G (lalla) -1.0°C 0.9°C before you leave lab! van't Hoff factor values (see directions on p. 3-6, steps 4-5): Solution D Solute Compound Naci Molality 0.505 mol/kg Ideal van't...
Molar mass determination by depression of freezing point lab I'm
stuck on calculating the moles of solute.. How do I calculate it?
Also can u please check if I've done everything else correctly..
The data I collected: Measured freezing point of pure water: 0.0
degrees Celsius Actual mass of solute used: 10.12g Freezing point
of solution (observed): -3.4 Celsius Mass of solution: 84.7g
Freezing point of a Solution of liqud unknown Freezing point depression: Trial #1. 0.0℃ (-3.4°C)= 3.4℃ Molality...
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