trial 1 and trial 2 Determine the delta tf (change in temperature) for the two t

Question

trial 1 and trial 2
Determine the delta tf (change in temperature) for the two trials performed.
Can you please do a step by step, for each question. I having special trouble with the last two questions. n Number: Data Table I Trial 2 Trial 1 121 ,59 LIA.qg Mass of empty test tube Mass of test tube + CH COOH Mass of CH COOH Freezing temperature of pure Co O s 21. 1 5,59 CH3COOH Mass of unknown Freezing temperature of solution | 7.2°C 7.5 Data Analysis/Calculations Use the following questions to assist in de show all calculations. Include the proper significant figures and units! termining the molecular mass of the unknown. Be sure to 1, Determine the T, for the two trials performed. 2. Using the Ky for acetic acid, determine the molality of the solutions for the two trials performed using the equation for freezing point depression in the introduction. Covert the mass of acetic acid from g to kg for each trial. Use the mass of acetic acid in each solution and the molality of the solution calculated in #2 to 3. 4. determine the number of mol of unknown present in each solution. Determine the molecular weight of the unknown using the mass weighed out for each trial Calculate the average molecular weight of the unknown based on the two trials. 5. 6.

Explanation / Answer

delta Tf = - Kf x m [negative because it is depression in temperature]

where,

m = molality = moles of solute / kg of solvent

moles = g/molar mass

Kf = 3.90 C/m [From literature for CH3COOH]

1. Determination of delta Tf

Trial 1

delta Tf = T(final) - T(initial) = 9.2 - 15.7 = -6.5 oC

Trial 2

delta Tf = T(final) - T(initial) = 9.5 - 10.4 = -0.9 oC

2. Determination of molality

Trial 1

- 6.5 = -3.90 x m

m = 1.67 molal

Trial 2

- 0.9 = -3.90 x m

m = 0.23 molal

3. Convert g of CH3COOH to kg of CH3COOH

Trial 1

15.6 g = 0.0156 kg

Trial 2

21.5 g = 0.0215 kg

4. Determine moles of unknown

Trial 1

m = 1.67 = moles/kg of solvent

moles = 1.67 mols x 0.0156 kg = 0.026 mols

Trial 2

m = 0.23 molal

moles = 0.23 x 0.0215 = 0.005 mols

5. Determine molar mass of unknown

Trial 1

moles = 0.026 mols = g/molar mass

molar mass = g/0.026 = 1.3/0.026 = 50 g/mol

Trial 2

moles = 0.005 mols = 1.3/molar mass

molar mass = 1.3/0.005 = 260 g/mol

6. Determine average molar mass

Average molar mass = 50 g/mol + 260 g/mol / 2 = 155 g/mol

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