Homework Answers
Ans 3). We calculate molarity of acetic acid in vinegar using equation
MNaOH VNaOH = Macetic acid Vacetic acid
MNaOH = 0.5167 ; Vacetic acid = 10mL
Macetic acid = (MNaOH x VNaOH )/ Vacetic acid = 0.05167 x VNaOH
| Trial | VNaOH | Macetic acid = 0.05167 x VNaOH |
| 1 | 13.30 | 0.6872 M |
| 2 | 13.55 | 0.7001 M |
| 3 | 13.51 | 0.6981 M |
Ans 4) Average molarity = (0.6872+0.7001+0.6981)/3 = 0.6951 M
Ans 5) Average molarity = 0.6951 M implys 0.6951 moles of acetic acid is present in 1000 ml of vinegar solution .
Mass of acetic acid present in 1000ml of solution = 0.6951 x 60 = 41.71 g
Density of vinegar = 1.05 g/ml ( From reference)
Mass of solution = density x volume = 1.05 x 1000 = 1050 g
Mass % = (mass of acetic acid / mass of solution)x100=(41.71x100)/1050) = 3.97%
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3. Calculate the molarity of the acetic acid in the vinegar solution for each trial. 4....
(TITRATION) What would be the calculations based on the trial data gathered? I obtained 50mL of...
(TITRATION) What would be the calculations based on the trial
data gathered?
I obtained 50mL of NaOH solution and then pipetted 5.0mL of
pickle juice into an Erlenmeyer flask. I then added 50mL of
distilled water and 2 drops of indicator.
Molarity of NaOH: 0.1001M Trial 1 Vinitial: 24.50 mL Vfinal: 45.72 mL Trial 2 Vinitial: 2.42 mL Vfinal: 23.74 mL Trial 3 Vinitial: 23.74 mL Vfinal: 44.22 mL Calculations Assume that the acid present in your food sample is...
Calculate the moles of acetic acid, molarity of the vinegar solution, and mass %. of acetic...
Calculate the moles of acetic acid, molarity of the vinegar solution, and mass %. of acetic acid in the vinegar using the average of three good trials used to titrate 5.00 mL of the vinegar. You may assume that the density of your vinegar sample is 1.01 g mL^-1. Drain off unused NaOH solution into a clean container and place it into a designated container. Wash the buret two times with regular tap water, then rinse with deionized water, and...
4. Vinegar is a solution of acetic acid. Acetic acid has a 1:1 ratio when reacted...
4. Vinegar is a solution of acetic acid. Acetic acid has a 1:1 ratio when reacted with NaOH. a. Suppose that the molarity of acetic acid in vinegar is 0.5 M, how many grams of acetic acid are present in the 5 ml of vinegar solution used in the titration? (molar mass of acetic acid = 60 g/mol). b. If vinegar has a density of 1.01 g/mol. Calculate the % of acetic acid in vinegar by mass. 5. 50 ml...
5. Determine the average value for the molarity of the acetic acid in vinegar. Show your...
5. Determine the average value for the molarity of the acetic acid in vinegar. Show your calculation 6. Determine the deviation between each individual molarity for acetic acid and the average value from question 4, above. Notice that the formula below uses absolute values: you only care about how big the difference is not whether cach trial value was higher or lower than the average. Deviation = (individual trial value)-(average value from #5) Trial Ideviation Trial 2 deviation Trial 3...
3. The manufacturer of the vinegar used in this experiment claims that the vinegar contains 5% acetic acid by weig...
3. The manufacturer of the vinegar used in this experiment claims that the vinegar contains 5% acetic acid by weight. Using your results, molarity of vinegar, and 1.0 g/ml for a density of vinegar, calculate the % acetic acid by weight. Can you tell if the manufacturer's claims is reasonable? Date Concentration of Acetic Acid in Vinegar: Results Instructor Name Partners Section Data molarity of the NaOH solution, M 0.250 M initial buret reading, ml - 0 ml 32mL Omt...
B. Analysis of Vinegar (Acetie Acid) Trial aceti ic aci S Advertised % Acetic Acid Trial...
B. Analysis of Vinegar (Acetie Acid) Trial aceti ic aci S Advertised % Acetic Acid Trial #2 Trial #3 NAT of vi 8. Volume of vinegar 10 ml 10ml 1.0 ml 9. Final buret reading 6.4 _ml 10. Initial buret reading 0.00 mL 6.4 ml. 13.4_mL 6.4 ml 7.0 ml 20.2__ml 13.4 ml 6.8 mi 11. Volume of NaOH 12. Molarity of vinegar (Show set-up) - M - 13. Average molarity of vinegar 14. Percent acetic acid in vinegar (Show...
Vinegar is a solution of acetic acid. Acetic acid has a 1:1 ratio when reacted with...
Vinegar is a solution of acetic acid. Acetic acid has a 1:1 ratio when reacted with NaOH. a. Suppose that the molarity of acetic acid in vinegar is 0.5 M, how many grams of acetic acid are present in the 5 ml of vinegar solution used in the titration? (molar mass of acetic acid = 60 g/mol). b. If vinegar has a density of 1.01 g/mol. Calculate the % of acetic acid in vinegar by mass.
13. Analysis of Vinegar (Acetic Acid) Advertised % Acetic Acid QUESTIONS: 1 Trial #1 Trial #2 Tri...
12) questions
the Molarity of Vinegar for trial #1 and #2
13. Analysis of Vinegar (Acetic Acid) Advertised % Acetic Acid QUESTIONS: 1 Trial #1 Trial #2 Trial #3 ate too f 1 ution an tion. What vo 8. Volume of vinegar ulate th of the d a mag 9. Final buret reading15o mS2 mt0.40 10. Initial buret reading 1st sar trials c s with used c 11. Volume of NaOH 12. Molarity of vinegar (Show set-up) 2. If 38.71...
1. Volume of Vinegar used in each trial = 5.00 mL 2. Molarity of NaOH used...
1. Volume of Vinegar used in each trial = 5.00 mL 2. Molarity of NaOH used in each trial = 0.20M 3. Volume of NaOH used in trail#1 (Final – Initial buret reading) = __17.80______ mL 4. Volume of NaOH used in trail#2 (Final – Initial buret reading) = ___18.20_____ mL 5. Average Volume of NaOH used = (#3 + #4) / 2 = ___18______ mL 6. Calculate Molarity of Acetic acid in vinegar = _________ M ?????? (molarity of...
Exp 9 titration of vinegar procedure b molarity of NaOH = .0859 ROCEDURE B. DETERMINATION OF...
Exp 9 titration of vinegar procedure b
molarity of NaOH = .0859
ROCEDURE B. DETERMINATION OF CONCENTRATION OF ACETIC ACID IN UNKNOWN SAMPLES Titration of vinegar solutions Trial 1 Trial 2 Trial 3 1. Volume of vinegar solution being titrated in milliliters (mL) 25.00 mL 25.00 mL 25.00 ml 2. Volume of vinegar solution being titrated in Liters (L) 0.0250 L 0.0250 L 0.0250 L 3. Final buret reading in ml 16. 20m 16.316L 16.25ml 4. Initial buret reading in...
(TITRATION) What would be the calculations based on the trial
data gathered?
I obtained 50mL of NaOH solution and then pipetted 5.0mL of
pickle juice into an Erlenmeyer flask. I then added 50mL of
distilled water and 2 drops of indicator.
Molarity of NaOH: 0.1001M Trial 1 Vinitial: 24.50 mL Vfinal: 45.72 mL Trial 2 Vinitial: 2.42 mL Vfinal: 23.74 mL Trial 3 Vinitial: 23.74 mL Vfinal: 44.22 mL Calculations Assume that the acid present in your food sample is...
Calculate the moles of acetic acid, molarity of the vinegar solution, and mass %. of acetic acid in the vinegar using the average of three good trials used to titrate 5.00 mL of the vinegar. You may assume that the density of your vinegar sample is 1.01 g mL^-1. Drain off unused NaOH solution into a clean container and place it into a designated container. Wash the buret two times with regular tap water, then rinse with deionized water, and...
4. Vinegar is a solution of acetic acid. Acetic acid has a 1:1 ratio when reacted with NaOH. a. Suppose that the molarity of acetic acid in vinegar is 0.5 M, how many grams of acetic acid are present in the 5 ml of vinegar solution used in the titration? (molar mass of acetic acid = 60 g/mol). b. If vinegar has a density of 1.01 g/mol. Calculate the % of acetic acid in vinegar by mass. 5. 50 ml...
5. Determine the average value for the molarity of the acetic acid in vinegar. Show your calculation 6. Determine the deviation between each individual molarity for acetic acid and the average value from question 4, above. Notice that the formula below uses absolute values: you only care about how big the difference is not whether cach trial value was higher or lower than the average. Deviation = (individual trial value)-(average value from #5) Trial Ideviation Trial 2 deviation Trial 3...
3. The manufacturer of the vinegar used in this experiment claims that the vinegar contains 5% acetic acid by weight. Using your results, molarity of vinegar, and 1.0 g/ml for a density of vinegar, calculate the % acetic acid by weight. Can you tell if the manufacturer's claims is reasonable? Date Concentration of Acetic Acid in Vinegar: Results Instructor Name Partners Section Data molarity of the NaOH solution, M 0.250 M initial buret reading, ml - 0 ml 32mL Omt...
B. Analysis of Vinegar (Acetie Acid) Trial aceti ic aci S Advertised % Acetic Acid Trial #2 Trial #3 NAT of vi 8. Volume of vinegar 10 ml 10ml 1.0 ml 9. Final buret reading 6.4 _ml 10. Initial buret reading 0.00 mL 6.4 ml. 13.4_mL 6.4 ml 7.0 ml 20.2__ml 13.4 ml 6.8 mi 11. Volume of NaOH 12. Molarity of vinegar (Show set-up) - M - 13. Average molarity of vinegar 14. Percent acetic acid in vinegar (Show...
Vinegar is a solution of acetic acid. Acetic acid has a 1:1 ratio when reacted with NaOH. a. Suppose that the molarity of acetic acid in vinegar is 0.5 M, how many grams of acetic acid are present in the 5 ml of vinegar solution used in the titration? (molar mass of acetic acid = 60 g/mol). b. If vinegar has a density of 1.01 g/mol. Calculate the % of acetic acid in vinegar by mass.
12) questions
the Molarity of Vinegar for trial #1 and #2
13. Analysis of Vinegar (Acetic Acid) Advertised % Acetic Acid QUESTIONS: 1 Trial #1 Trial #2 Trial #3 ate too f 1 ution an tion. What vo 8. Volume of vinegar ulate th of the d a mag 9. Final buret reading15o mS2 mt0.40 10. Initial buret reading 1st sar trials c s with used c 11. Volume of NaOH 12. Molarity of vinegar (Show set-up) 2. If 38.71...
Exp 9 titration of vinegar procedure b
molarity of NaOH = .0859
ROCEDURE B. DETERMINATION OF CONCENTRATION OF ACETIC ACID IN UNKNOWN SAMPLES Titration of vinegar solutions Trial 1 Trial 2 Trial 3 1. Volume of vinegar solution being titrated in milliliters (mL) 25.00 mL 25.00 mL 25.00 ml 2. Volume of vinegar solution being titrated in Liters (L) 0.0250 L 0.0250 L 0.0250 L 3. Final buret reading in ml 16. 20m 16.316L 16.25ml 4. Initial buret reading in...
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