Chemistry Homework Solutions
Problem
#54113

COLLIGATIVE PROPERTIES

1. I need help with the standard deviation/percentage error in the attached lab experiment.

2. Also need help graphing for the ethyl acelate and unknown solvent -- plotting negative delta T versus the total volume of solute that was added (2 graphs) and calculation of the slopes

3.  I am not certain but I believe this question relates to the accepted value of the freezing point of cyclohexane 6.5 degrees and a Kf of 20.0 (not the 8.0 degrees that was found in the lab).  Can an answer to the following question be determined using this assumption?  Relationship of freezing point depression constant to its enthalpy of fusion and its entropy of fusion with the following formulas

delta H fus = MRT(squared)
                   ____________

                    1000Kf

and delta S fus = MRT of cyclohexane
                       ____________

                       1000 Kf

t is the freezing point of pure cyclohexane and R is the gas constant.  Units need to be arranged so that delta H comes out in KJ mol-1 and delta s comes out in JK-1 mol-1.  

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Colligative Properties.doc  View File

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Colligative Properties.doc
Run Volume of ethyl acetate Moles of ethyl acetate Molality (moles/kg)
(Tf (K) Kf (K molal-1)

1 0.20 mL 2.03 Ч 10-3 g 0.10433 3.333 31.9

2 0.40 mL 4.06 Ч 10-3 g 0.20866 5.667 27.2

3 0.60 mL 6.10 Ч 10-3 g 0.31298 9.667 30.9





Average Kf 30.0



Sample calculation for Run 1:

Volume of ethyl acetate = 0.20 mL



Molality = 2.03 Ч 10-3 mole / 0.019475 kg

= 0.10433 molal

Average Tf = (4 + 5 + 5)/3 = 4.667 oC

(Tf = 8 – 4.667 = 3.333 oC = 3.333 K



Same calculation is repeated for the other two runs.

The average Kf is found to be 30.0 K molal-1.

Run Volume of unknown (Tf

(K) molal Moles of unknown Mass of unknown Molar mass (g/mol)

1 0.20 mL 0.5 0.01667 3.246 Ч 10-4 0.26 g 800.98

2 0.40 mL 1 0.03333 6.49 Ч 10-4 0.52 g 801.23

3 0.60 mL 2 0.06667 1.298 Ч 10-3 0.78 g 600.92





Average molar mass 734.4



Sample calculation for Run 1:



moles of unknown = molal Ч kg of solvent

= 0.01667 moles / kg Ч 0.019475 kg

= 3.246 Ч 10-4 moles



The unknown molar mass is 734.4 g mol-1.

(All (Tf values are shown in the table).

Solution Summary

The solution provides a detailed and step-by-step explanation for the problem. It includes a 3-page Word file and calculations/graphing in Excel.

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