Question

Here's the data-

the exact concentration of NaOH was 0.100M
the vinegar volume (mL) for trial 1, trial 2, and trial 3 were all 5.00
the final burette reading of NaOH (mL) for trial 1 was 43.75, for trial 2 it was 42.80, and for trial 3 it was 43.20
the initial burette reading of NaOH (mL) for trial one was 0.90, for trial 2 it was 0.20, for trial 3 it was 0.50
1. average volume of NaOH in liters (L)
2. average moles of NaOH (mol NaOH)
3. average moles of acidic acid (mol CH3COOH)
4. average molarity of acetic acid (M)
5. average mass of acidic acid (g CH3COOH)
6. average mass of vinegar (g)
7.%(m/m) CH3COOH in vinegar

Answer 1

To calculate the average volume of NaOH in liters, subtract the initial burette reading from the final burette reading for each trial and divide by the number of trials. Use the formula Moles = Molarity x Volume (in liters) to find the average moles of NaOH. The average moles of acetic acid (CH3COOH) are equal to the average moles of NaOH.

Step-by-step explanation:

To calculate the average volume of NaOH in liters, we need to add the final burette reading and the initial burette reading for each trial, and then divide by the number of trials. For trial 1, the volume of NaOH is (43.75 mL - 0.90 mL) = 42.85 mL. Converting this to liters, we get 0.04285 L. Repeat this calculation for the other trials and then calculate the average.

To find the average moles of NaOH, we can use the formula Moles = Molarity x Volume (in liters). Multiply the molarity of NaOH (0.100 M) by the average volume (in liters) calculated in the previous step to get the average moles of NaOH.

To find the average moles of acetic acid (CH3COOH), we use the balanced chemical equation:

NaOH + CH3COOH → NaCH3COO + H2O

We know that the moles of NaOH and CH3COOH are in a 1:1 ratio. Therefore, the average moles of acetic acid are equal to the average moles of NaOH.