PART A. FREEZING POINT OF LAURIC ACID Mass lauric acid, mass, Tf, trial 1 Tf, trial 2 1,2. Average Tf Data Unrounded Value S 43.5000 1.952 42.7 44.3 ✔ g

YOUR PROGRESS ON FREEZING POINT OF LAURIC ACID °C °C Rounded Value 43.5 Unit °C Unit PART B. FREEZING POINT OF SOLUTION OF (LAURIC ACID+ UNKNOWN) UNKNOWN ID AND MASS Unknown ID VB Mass solvent, mass, Mass solute, mass2 ✓ Table to report mass values from experiment • Kf of lauric acid is 3.90 °C kg mol-¹ • Use your average freezing point for lauric acid Data 1.952 3 g Note: report LabQuest values as is for intercepts and slopes. Unit g GRAPH DATA 3. Slope of line 1, m₁ 4. Intercept of line 1, b₁ Table to report graph data from LabQuest 5. Slope of line 2, m₂ 6. Intercept of line 2, b₂ CALCULATIONS 7.8. T 9,10. ΔΤ, 11,12. Molar mass of unknown Data D D Unrounded Value 5 Unit CE dill 2 2 2 2 Rounded Value 3 > Unit 2 2 X SCORE: 21% Choose molar mass closest to your value: ERROR 13,14. % Error, molar mass CORRECT YOUR PROGRESS ON FREEZING POINT OF SOLUTION OF (LAURIC ACID + UNKNOWN) 1 17 Unrounded Value SAVED: 5.16. T/ PART C. ADDITIONAL QUESTION 6 Unrounded Value POINTS AWARDED 3 53 Rounded Value AUTOSOLVED The molar mass of a-linolenic acid is 278.43 g mol-¹ Rounded Value 0 17 If 0.213 g of mass, (a-linolenic acid) is dissolved in 2.912 g of mass3 (lauric acid), calculate the expected freezing point of the solution. NOT FINISHED 21 23 If 0.213 g of mass (a-linolenic acid) is dissolved in 2.912 g of mass, (lauric acid), calculate the expected freezing point of the solution. 15,16. T CORRECT YOUR PROGRESS ON THE ADDITIONAL QUESTION Unrounded Value Extra Stats to show: 02 YOUR OVERALL PROGRESS 7 25 CORRECT ANSWERS Units Rounded Value POINTS AWARDED 0 10 Calculations POINTS AWARDED 16 79 AUTOSOLVED Rounding Concepts AUTOSOLVED 0 25 02 NOT FINISHED NOT FINISHED 23 34 2/2/nman I Part A Mass of lauric acid 5₂ 1.952 After freezing Value of y = Tf₁ = 1-4.0914, g 1 299 27 Before freezing m₁ = Before freezing 48.96 aftur freezing = -0.22788 Locker/ Desk No. Freezing point of lauric acid: Trial 1: 42.7°C (42.7° 52.9) x2 = 47.8°c (326.95k) Trial 2: 52.9°C 2.099 Average freezing point of pure lauric acid, Te 320.95 K Part B Unknown ID A mass of unknown 0.315g Course & Section No. Tr Trials Before freezing m₁ = -8.0645m₁ Before freeding b₁ = 47.151 After freezing m2=0.5446 | 'value of y = Tf' = b2 = 37.599 Average value of freezing & point of lauric acid/unknown Solution Tf= Show cate how you calculated the value of te for one of your trials: Depression of freezing point, Stp = Te-Tp. M