Question

UUTUUDIY Report Titration Run 1 Run 3 Name Laura A cense Run 2 Volume of Vinegar Used o radac Concentration of NaOH 0.100 M 0.100 M 0.100 M 1.45 mL 1 1 19.68 ml 107.69 m2 Final Buret reading 19.68 m 36.78me lauo some Volume of NaOH Used 6.01sas 0.01710 0,ole 871 (Final Buret reading - 18.a I rolML 18h 16.87 Initial Buret reading) Calculations: 1. Calculate the number moles of base required to reach the endpoint for each run. 2. Calculate the number of moles of acid in each sample. 3. Using your answer to calculation 2, calculate the molarity of the vinegar sample for each run. 4. Calculate the average molarity of the vinegar. 5. The true concentration of the vinegar is 5.00g/100mL of vinegar. Calculate the true molarity of the vinegar. 6. Calculate the percent error using your answers to calculations 4 and 5.

question 6

Answer 1

Answer: 6

The calculated average molarities from the titration = (0.9125 + 0.8550 + 0.8435)/3 = 0.8703 M

The molar mass of CH₃COOH (Vinegar) = 60.05 g/mol (Acid present as Vinegar in True solution)

Now, the moles of CH₃COOH (Vinegar) = 5/60.05 = 0.08326 moles

Total Volume of True solution = 0.1 L

and Molarity of True solution = 0.08326/0.1 = 0.8326 M

Now % error = (|experiment - True|/True ) x 100 = 4.52 %

Please let me know, if you have any doubt by commenting below the answer.

Thanks