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Answer the last part please: RECORD ALL DATA IN INK. SHoW COMPLETE \"SET-UPS\" (

ID: 561602 • Letter: A

Question

Answer the last part please:

RECORD ALL DATA IN INK. SHoW COMPLETE "SET-UPS" (and the rounding off) FOR ALL CALCULATIONS on a separate sheet. Use the correct number of significant figures and correct units when recording data and when calculating the results. Part I Freezing Point of Pure PDB. 12 1. Mass of test tube PDB 2. Mass of test tube 3. Mass of PDB 4. Freezing point of PDB, Trial 1A [from graph] 5. Freezing point of PDB, Trial IB [from graph] 6. Average Freezing point of PDB 42 2010 30 025a 53.00 53.5C S3.00+ Su.00 10-53.5 Freezing Point of a Solution of an Unknown Solid in PDB Part II. 1. Mass of sample vial + solute sample Unknown solute sample number: 2. Mass of sample vial +solute sample after 1st transfer 3. Freezing point of solution, Trial 2A Ifrom graph

Explanation / Answer

1.

Trial #2

Freezing Point change = 53.5 – 49 C = 4.5 C

Trial #3

Freezing Point change = 53.5 – 44.25 C =9.25 C

2.

Trial #2

Mass of solute in solution = 11.029 – 10.033 g = 0.996 g

Trial #3

Mass of solute in solution = 10.033 – 7.61 g =2.423 g

3.

Trial #2 & Trial # 3

Mass of PDB in solution = 30.035 g (Data missing, use the amount present in the experiment if not same as in trial #1)

4.

Molality, m = moles/ kg solvent

= moles of solute / 30.035 * 1000 = 33.3 moles of solute

Tf = Kf * m

Trial #2

4.5 C = 7.1 C/m * 33.3 * moles of solute

Moles of solute = 4.5 / (7.1 * 33.3) = 0.019

Trial #3

9.25 C = 7.1 C/m * 33.3 * moles of solute

Moles of solute = 0.039

5.

Trial #2

Molar Mass of solute = Mass / Moles

= 0.996 / 0.019 = 52.3 g/mol

Trial #3

Molar Mass of solute = Mass / Moles

= 2.423 / 0.039 = 61.9 g/mol

6.

Average Molar Mass of solute

= (52.3 + 61.9)/2 = 57.1 g/mol

7.

Average deviation = ((61.9 – 57.1) + (57.1 – 52.3))/2

= 4.8

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