Sure, let's go through the computation step-by-step to determine the number of moles of NaOH for each trial.
First, recall the following key points:
- The molarity of the solution (M) is the number of moles of solute per liter of solution.
- The volume needs to be in liters to use the molarity properly.
Here are the steps we'll take for each trial:
### Conversion of Volume to Liters
The initial volume is given in milliliters (mL). We need to convert it to liters (L) for our calculations by dividing by 1000.
### Multiplying by Molarity
Next, we multiply the volume in liters by the molarity to find the number of moles of NaOH. We'll round the final answer to four significant figures.
### Trial 1
1. The volume is \( 12.49 \, \text{mL} \).
2. Convert this volume to liters:
[tex]\[
12.49 \, \text{mL} = \frac{12.49}{1000} \, \text{L} = 0.01249 \, \text{L}
\][/tex]
3. Multiply by the molarity (which we'll assume is \( 1.0 \, \text{mol/L} \) for the purpose of this calculation):
[tex]\[
\text{moles of NaOH} = 0.01249 \, \text{L} \times 1.0 \, \text{mol/L} = 0.01249 \, \text{mol}
\][/tex]
4. Round to four significant figures:
[tex]\[
\text{moles of NaOH} = 0.01249 \approx 0.0125 \, \text{mol}
\][/tex]
### Trial 2
1. The volume is \( 12.32 \, \text{mL} \).
2. Convert this volume to liters:
[tex]\[
12.32 \, \text{mL} = \frac{12.32}{1000} \, \text{L} = 0.01232 \, \text{L}
\][/tex]
3. Multiply by the molarity:
[tex]\[
\text{moles of NaOH} = 0.01232 \, \text{L} \times 1.0 \, \text{mol/L} = 0.01232 \, \text{mol}
\][/tex]
4. Round to four significant figures:
[tex]\[
\text{moles of NaOH} = 0.01232 \approx 0.0123 \, \text{mol}
\][/tex]
### Trial 3
1. The volume is \( 11.87 \, \text{mL} \).
2. Convert this volume to liters:
[tex]\[
11.87 \, \text{mL} = \frac{11.87}{1000} \, \text{L} = 0.01187 \, \text{L}
\][/tex]
3. Multiply by the molarity:
[tex]\[
\text{moles of NaOH} = 0.01187 \, \text{L} \times 1.0 \, \text{mol/L} = 0.01187 \, \text{mol}
\][/tex]
4. Round to four significant figures:
[tex]\[
\text{moles of NaOH} = 0.01187 \approx 0.0119 \, \text{mol}
\][/tex]
### Summary of Results
- Trial 1: \( 12.49 \, \text{mL} \) = \( 0.0125 \, \text{mol NaOH} \)
- Trial 2: \( 12.32 \, \text{mL} \) = \( 0.0123 \, \text{mol NaOH} \)
- Trial 3: \( 11.87 \, \text{mL} \) = \( 0.0119 \, \text{mol NaOH} \)
These are the number of moles of NaOH in each trial using the provided volumes and keeping the results to four significant figures.
