From a titration experiment the following results were taken: trial 1- initial volume = 0 and final volume = 8.5 mL. Trial 2- initial reading 8.5 mL and final reading 14.2 mL. Trial 3- initial reading 14.2 mL and ginal reading 24.8 mL. Thus from the results select two concordant titres and calculate the two mean titres, molarity, overall maximum error and overall % error.

Answer 1

To analyze the results from your titration experiment, let's first summarize the data you've provided:

1. Trial 1:
   Initial Volume: 0 mL
   Final Volume: 8.5 mL
   Titre: \(8.5 - 0 = 8.5\) mL

2. Trial 2:
   Initial Volume: 8.5 mL
   Final Volume: 14.2 mL
   Titre: \(14.2 - 8.5 = 5.7\) mL

3. Trial 3:
   Initial Volume: 14.2 mL
   Final Volume: 24.8 mL
   Titre: \(24.8 - 14.2 = 10.6\) mL

Next, we'll identify the concordant titres. Concordant titres are those that are very close to each other (typically within 0.1 mL). In this case, we have:

• Titre 1: \(8.5\) mL
• Titre 2: \(5.7\) mL
• Titre 3: \(10.6\) mL

Since there are no values that are extremely close, we can use two of the trials with some common practice as follows. However, let's select two values that have the minimal differences.

Concordant Titre Selection

Since we do not have any concordant titres as per the strict definition of concordance, we can treat "8.5" and "5.7" as part of our analysis for calculation purposes but for accuracy consider them distinct. However for calculation let's go with trials that give us near ranges manually as we have only 3 trials.

From the calculations, we will assume we take \(8.5\) mL and probably average all different even if they are not concordant.

Mean Titre Calculation

For the sake of calculating averages, we will take all values (8.5, 5.7, 10.6) but since they are too distinct could lead to incomparable averages.

• Mean Titre = \((8.5 + 5.7 + 10.6) / 3 = 8.26 , \text{mL} \)

Molarity Calculation

To calculate the molarity of a solution, the equation we use is:

\[ \text{Molarity} (M) = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]

To proceed with this calculation, you'll need the following:

1. Moles of Solute: This typically requires the mass of the solute and its molar mass.
2. Volume of Titration Solution: We will take the mean titre of \(8.26\) mL which is \(0.00826\) L.

Without specific solute mass or molarity, further numerical calculation for molarity cannot be completed.

Overall Maximum Error Calculation

Assuming the uncertainty in the reading is ±0.1 mL (which is common in such experiments):

• For the three trials, maximum absolute error = ±0.1 mL
• Therefore, the overall maximum error for the mean titre calculation would also be considered as ±0.1 mL.

Overall % Error Calculation

The overall % error is calculated as: \[ \text{Overall % Error} = \left( \frac{\text{Maximum Error}}{\text{Measured Value}} \right) \times 100 \]

Using the average titre of \(8.26\) mL:

\[ \text{Overall % Error} = \left( \frac{0.1 , \text{mL}}{8.26 , \text{mL}} \right) \times 100 \approx 1.21% \]

Summary of Results

• Mean Titre: \(8.26 , \text{mL}\) (average of all trials)
• Molarity: Not calculable without additional info
• Overall Maximum Error: ±0.1 mL
• Overall % Error: Approximately \(1.21%\)

If you have further information pertaining to the solute or additional corrections, let me know!