Answer :
To determine how much water must be added to obtain a 1.00 L solution with a final concentration of 4.00 M from an initial concentration of 15.9 M, we can use the dilution formula:
[tex]\[ C_1 \times V_1 = C_2 \times V_2 \][/tex]
where:
- [tex]\( C_1 \)[/tex] is the initial concentration (15.9 M)
- [tex]\( V_1 \)[/tex] is the initial volume
- [tex]\( C_2 \)[/tex] is the final concentration (4.00 M)
- [tex]\( V_2 \)[/tex] is the final volume (1.00 L)
First, we need to solve for [tex]\( V_1 \)[/tex], which is the volume of the concentrated nitric acid solution required.
Rearranging the dilution formula, we get:
[tex]\[ V_1 = \frac{C_2 \times V_2}{C_1} \][/tex]
Substitute the given values into the equation:
[tex]\[ V_1 = \frac{4.00 \, \text{M} \times 1.00 \, \text{L}}{15.9 \, \text{M}} \][/tex]
[tex]\[ V_1 = \frac{4.00}{15.9} \][/tex]
[tex]\[ V_1 \approx 0.2516 \, \text{L} \][/tex]
So, [tex]\( V_1 \approx 0.2516 \)[/tex] L of the 15.9 M nitric acid is required.
Next, we need to determine the volume of water that needs to be added. Since the final volume [tex]\( V_2 \)[/tex] is 1.00 L and the volume of the concentrated solution is 0.2516 L, the volume of water added can be found by subtracting [tex]\( V_1 \)[/tex] from [tex]\( V_2 \)[/tex]:
[tex]\[ \text{Volume of water added} = V_2 - V_1 \][/tex]
[tex]\[ \text{Volume of water added} = 1.00 \, \text{L} - 0.2516 \, \text{L} \][/tex]
[tex]\[ \text{Volume of water added} \approx 0.7484 \, \text{L} \][/tex]
Therefore, approximately 0.7484 liters (or 748.4 mL) of water must be added to the concentrated nitric acid to obtain 1.00 L of a 4.00 M solution.
[tex]\[ C_1 \times V_1 = C_2 \times V_2 \][/tex]
where:
- [tex]\( C_1 \)[/tex] is the initial concentration (15.9 M)
- [tex]\( V_1 \)[/tex] is the initial volume
- [tex]\( C_2 \)[/tex] is the final concentration (4.00 M)
- [tex]\( V_2 \)[/tex] is the final volume (1.00 L)
First, we need to solve for [tex]\( V_1 \)[/tex], which is the volume of the concentrated nitric acid solution required.
Rearranging the dilution formula, we get:
[tex]\[ V_1 = \frac{C_2 \times V_2}{C_1} \][/tex]
Substitute the given values into the equation:
[tex]\[ V_1 = \frac{4.00 \, \text{M} \times 1.00 \, \text{L}}{15.9 \, \text{M}} \][/tex]
[tex]\[ V_1 = \frac{4.00}{15.9} \][/tex]
[tex]\[ V_1 \approx 0.2516 \, \text{L} \][/tex]
So, [tex]\( V_1 \approx 0.2516 \)[/tex] L of the 15.9 M nitric acid is required.
Next, we need to determine the volume of water that needs to be added. Since the final volume [tex]\( V_2 \)[/tex] is 1.00 L and the volume of the concentrated solution is 0.2516 L, the volume of water added can be found by subtracting [tex]\( V_1 \)[/tex] from [tex]\( V_2 \)[/tex]:
[tex]\[ \text{Volume of water added} = V_2 - V_1 \][/tex]
[tex]\[ \text{Volume of water added} = 1.00 \, \text{L} - 0.2516 \, \text{L} \][/tex]
[tex]\[ \text{Volume of water added} \approx 0.7484 \, \text{L} \][/tex]
Therefore, approximately 0.7484 liters (or 748.4 mL) of water must be added to the concentrated nitric acid to obtain 1.00 L of a 4.00 M solution.
