To compute the concentration (molarity) of HCl for each of the three trials, we'll follow a few steps. We know that the formula for molarity is given by:
[tex]\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \][/tex]
Given:
- The number of moles of HCl is 1 mol for each trial.
We also need to convert the volume from milliliters (mL) to liters (L) by dividing by 1000.
### Trial 1
Volume: 20.61 mL
[tex]\[ \text{Volume in liters} = \frac{20.61}{1000} = 0.02061 \text{ L} \][/tex]
Molarity:
[tex]\[ \text{Molarity} = \frac{1 \text{ mol}}{0.02061 \text{ L}} \][/tex]
[tex]\[ \text{Molarity Trial 1} = 48.52 \text{ M} \][/tex]
### Trial 2
Volume: 20.06 mL
[tex]\[ \text{Volume in liters} = \frac{20.06}{1000} = 0.02006 \text{ L} \][/tex]
Molarity:
[tex]\[ \text{Molarity} = \frac{1 \text{ mol}}{0.02006 \text{ L}} \][/tex]
[tex]\[ \text{Molarity Trial 2} = 49.85 \text{ M} \][/tex]
### Trial 3
Volume: 19.67 mL
[tex]\[ \text{Volume in liters} = \frac{19.67}{1000} = 0.01967 \text{ L} \][/tex]
Molarity:
[tex]\[ \text{Molarity} = \frac{1 \text{ mol}}{0.01967 \text{ L}} \][/tex]
[tex]\[ \text{Molarity Trial 3} = 50.84 \text{ M} \][/tex]
Hence, the concentration of HCl for each trial is:
- Trial 1: [tex]\( 48.52 \text{ M} \)[/tex]
- Trial 2: [tex]\( 49.85 \text{ M} \)[/tex]
- Trial 3: [tex]\( 50.84 \text{ M} \)[/tex]
These results represent the concentrations of HCl in each respective trial.
