To determine the concentration of HCl for each trial, we'll use the formula for molarity:
[tex]\[ \text{Molarity} (M) = \frac{\text{Number of moles of solute} (\text{mol})}{\text{Volume of solution} (\text{L})} \][/tex]
Given:
- The number of moles of HCl used in each trial = 0.01 mol
- The volumes of HCl used in each trial are given in milliliters and need to be converted to liters:
- Trial 1: 20.61 mL
- Trial 2: 20.06 mL
- Trial 3: 19.67 mL
First, let's convert the volumes from mL to L:
[tex]\[ \text{Volume (L)} = \text{Volume (mL)} \div 1000 \][/tex]
For Trial 1:
[tex]\[ \text{Volume of HCl} = 20.61 \text{ mL} \div 1000 = 0.02061 \text{ L} \][/tex]
For Trial 2:
[tex]\[ \text{Volume of HCl} = 20.06 \text{ mL} \div 1000 = 0.02006 \text{ L} \][/tex]
For Trial 3:
[tex]\[ \text{Volume of HCl} = 19.67 \text{ mL} \div 1000 = 0.01967 \text{ L} \][/tex]
Now, we can calculate the molarity for each trial:
### Trial 1
[tex]\[ \text{Molarity} = \frac{0.01 \text{ mol}}{0.02061 \text{ L}} \approx 0.4852 \text{ M} \][/tex]
### Trial 2
[tex]\[ \text{Molarity} = \frac{0.01 \text{ mol}}{0.02006 \text{ L}} \approx 0.4985 \text{ M} \][/tex]
### Trial 3
[tex]\[ \text{Molarity} = \frac{0.01 \text{ mol}}{0.01967 \text{ L}} \approx 0.5084 \text{ M} \][/tex]
Therefore, the concentrations of HCl for the three trials are:
- Trial 1: 0.4852 M
- Trial 2: 0.4985 M
- Trial 3: 0.5084 M
