Determination of a Molar Mass by Freezing Point Depression: This experiment used Stearic Acid Experiments done in many first-year chemistry courses. They often use cyclohexane as the solvent and compounds such as p-dichlorobenzene, biphenyl, and naphthalene as the unknown solutes. Describe the advantages of this experiment over those.
Please answer in words and not equations, what are the advantages over the others?
Solution:
The following advantages of stearic acid during molar mass determination.
>>The determination of molar mass by stearic acid is considered as the less expensive because stearic acid is inexpensive fatty acid.
>> Stearic acid is less toxic fatty acid, hence the volume of waste can be reduced to zero if we use stearic acid as starting materials for other laboratory experiments.
>> It is considered as green chemistry experiment due to less expensive and less hazardous nature of stearic acid.
Determination of a Molar Mass by Freezing Point Depression: This experiment used Stearic Acid Experiments done...
Molar Mass Determination by Freezing Point Depression Calculate and enter the freezing point depression of a solution of 57.6 g ethylene glycol (C2H602) in 734 g H20. Kffor H20 is 1.86 °C kg/mol. °C -2.53 1 homework pts Submit Answer Incorrect. Tries 3/5 Previous Tries A solution which contains 71.9 g of an unknown molecular compound in 363 g of water freezes at -3.85°C. What is the molar mass of the unknown? g/mol 1homework pts Submit Answer Tries 0/5 Molar...
Molar Mass by Freezing Point Depression con Pre-Laboratory Assignment I. The following data were obtained in an experiment designed to find the molar mass of a solute by freezing point depressio Solvent: para-dichlorobenzene Freezing point ofpure solvent: 53.02"C Mass of unknown substance: 2.04g Freezing point depression constant: 7.1°C/m Mass of para-dichlorobenzene: 24.80 g Freezing point of solution: 50.78 °C a. Determine the freezing point depression, Δ b. Using Equation 4, calculate the molar mass of the unknown substance. .2.au"-7. ↓...
In experiment 8 (Molar Mass Determination by Depression By Freezing Point) Why is the "molar mass" is calculated and not the "molecular weight?" In General Chem 2.
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....
A Molar Mass from freezing-point Depression. 1. Provide definitions for the following terms: a. Solution-cv b. Solute c. Solvent d. Colligative property e. Freezing-point depression a. What is the objective of this experiment? b. How will that objective be achieved? 3. A 0.2436-g sample of an unknown substances was dissolved in 20.0 mL of cyclohexane. The density of cyclohexane is 0.779 g/mL. The freezing-point depression was 2.5*C. Calculate the molar mass of the unknown substance. 4. What safety rules must...
Freezing Point Depression: Determination of the Molar Mass of an Unknown Substance Laboratory Record Unknown Identifier #4 (All unknowns are nonelectrolytes, soi-1) Determination of the freezing point of pure l-butyl alcohol: (Record time-temperature data on the next page) 29.62 1. Mass of empty test tube 2. Mass of test tube and t-butyl alcohol after determination of freezing point of pure t-butyl alcohol 50.25 20.71 3. Mass of t-butyl alcohol Determination of the molar mass of an unknown: (Record time-temperature data...
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...
Molar Mass Determination by Freezing Point Depression Calculate and enter the freezing point depression of a solution of 65.1 g ethylene glycol (C2H602) in 792 g H20. Kffor H2O is 1.86 °C kg/mol. °C 1homework pts Submit Answer Tries 0/5 A solution which contains 60.9 g of an unknown molecular compound in 325 g of water freezes at -3.24°C. What is the molar mass of the unknown? g/mol 1homework pts Submit Answer Tries 0/5
Determination of Molar Mass by freezing Point depression
Not sure how to answer these two questions
Calcium chloride, CaCI_2 is commonly used as road salt. If one mole of CaCI_2 is dissolved in 1 kg v of water, would the freezing point of the solution be greater than, less than, or the same as a solution prepared by dissolving one mole of table sugar (sucrose, C_12H_22O_11) in 1 kg of water? Explain your answer. A student followed our lab procedure...
Chemistry:
Molar mass determination and freezing point depression.
If your unknown solute sample weighed 0.634 grams, but some of it stuck to the test tube wall when you were pouring it in and did not dissolve in the t-butyl alcohol, what sort of error will this cause in the calculated molar mass of your unknown solute? Explain. What would be the freezing point of an aqueous solution that contains 10.3 grams of ethylene glycol [(C_2H_4(OH)_2] in 100 mL of water?