I need help! Please Results and Discussion - Experiment 3 - Acids an d Bases: An

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I need help! Please

Results and Discussion - Experiment 3 - Acids an d Bases: Analysis A. Analysis of vinegar 1. Unknown vinegar number 2. Mass of flask 3. Mass of flask+5.00 mL of vinegar 4. Mass of 5.00 mL of vinegar 5. Density of vinegar 6. Molarity of sodium hydroxide solution 7 10.70 Titrations flask 1 flask 2 flask 3 7. Final buret reading 8. Initial buret reading Leme amu )mL- Volume ofNdH soin. 10. Moles of NaOH 11. Moles of acetic acis00s 12. Molarity of acetic acid 13. Average molarity of acetic acid in vinegar Deviation of each molarity from average 14. 15. Average deviation of molarity (see Expt. 4 Mass percent of acetic acid in vinegar (show calculations below) 16. 0 2015 Cengage Learning. All Rights Reserved May not be scanned, capied or duplicated, or posted to a publicly accessible

Explanation / Answer

No of moles of acetic acid will be equal to no. of moles of NaOH reacted as one mole of acetic acid requires one mole of NaOH for titration

No. of moles of acetic acid = 0.00019 0.0002 0.0001 moles

Molarity of acetic acid in vinegar = No. of moles of acetic acid / Volume of vinegar ( 5 ml = 0.005 L)   

Molarity of acetic acid in vinegar = 0.038 0.04 0.02 mol/L

Average molarity of acetic acid in vinegar = (0.038 + 0.04 + 0.02)/3 = 0.033 mol/L   

Deviation of molarity from average= 0.005 0.007 -0.013

Average deviation = 0.0052

No. of moles acetic acid = Molarity * volume = 0.033 mol/L * 0.005 = 0.000165 moles

Molar mass of acetic acid = 60 g/mol

Mass of acetic acid = 0.0099 g = 9.9 mg

Mass percent of acetic acid in vinegar = (Mass of acetic acid / Mass of vinegar) * 100 % = (0.0099 / 10.76) * 100

Mass percent of acetic acid in vinegar = 0.092 % Answer

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