Calculate the final concentration of
a. 4.0 L of a 4.0 M HNO3 solution is added to water so that the final volume is 8.0L
b. Water is added to 0.35L of a 6.0M KOH solution to make 2.0L of a diluted KOH solution.
c.A 20.0 mL sample of 8.0% (m/v)NaOH is diluted with water so that the final volume is 200.0mL.
d. A 6.0mL sample of 50.0% (m/v) acetic acid HC2H3O2 solution is added to water to give a final volume of 25mL. pls help even just the formula on how to solve it please.
I think all of these can be done with c1v1 = c2v2
c = concn
v = volume
thanks DrBob222 i already figure out how to do it
To calculate the final concentration of a solution after dilution, we can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution
V2 = final volume of the solution
Now, let's calculate the final concentration for each given scenario:
a. 4.0 L of a 4.0 M HNO3 solution is added to water so that the final volume is 8.0L
Using the formula: C1V1 = C2V2
C1 = 4.0 M
V1 = 4.0 L
C2 = ?
V2 = 8.0 L
4.0 M * 4.0 L = C2 * 8.0 L
16 M·L = 8C2 M·L
Solving for C2: C2 = 16 M·L / 8.0 L = 2.0 M
The final concentration is 2.0 M.
b. Water is added to 0.35L of a 6.0M KOH solution to make 2.0L of a diluted KOH solution.
Using the formula: C1V1 = C2V2
C1 = 6.0 M
V1 = 0.35 L
C2 = ?
V2 = 2.0 L
6.0 M * 0.35 L = C2 * 2.0 L
2.10 M·L = 2C2 M·L
Solving for C2: C2 = 2.10 M·L / 2.0 L = 1.05 M
The final concentration is 1.05 M.
c. A 20.0 mL sample of 8.0% (m/v) NaOH is diluted with water so that the final volume is 200.0 mL.
To calculate the final concentration, we need to convert the percentage to molarity, assuming the density of water is approximately 1 g/mL.
8% (m/v) means 8 grams of NaOH in 100 mL of solution.
8 g / 100 mL = 0.08 g/mL
Now we calculate the molarity (M):
M = (mass of solute / molar mass of solute) / volume of solution
molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
M = (0.08 g / 40.00 g/mol) / 0.100 L = 0.20 M
Now, we can use the formula: C1V1 = C2V2
C1 = 0.20 M
V1 = 20.0 mL (convert to liters: 20.0 mL / 1000 = 0.020 L)
C2 = ?
V2 = 200.0 mL (convert to liters: 200.0 mL / 1000 = 0.200 L)
0.20 M * 0.020 L = C2 * 0.200 L
0.0040 M·L = 0.20C2 M·L
Solving for C2: C2 = 0.0040 M·L / 0.200 L = 0.02 M
The final concentration is 0.02 M.
d. A 6.0 mL sample of 50.0% (m/v) acetic acid HC2H3O2 solution is added to water to give a final volume of 25 mL.
To calculate the final concentration, we need to convert the percentage to molarity, assuming the density of water is approximately 1 g/mL.
50% (m/v) means 50 grams of acetic acid in 100 mL of solution.
50 g / 100 mL = 0.50 g/mL
Now we calculate the molarity (M):
M = (mass of solute / molar mass of solute) / volume of solution
molar mass of HC2H3O2 = 2(1.01 g/mol) + 12.01 g/mol + 3(1.01 g/mol) + 16.00 g/mol = 60.05 g/mol
M = (0.50 g / 60.05 g/mol) / 0.100 L = 0.83 M
Now, we can use the formula: C1V1 = C2V2
C1 = 0.83 M
V1 = 6.0 mL (convert to liters: 6.0 mL / 1000 = 0.006 L)
C2 = ?
V2 = 25 mL (convert to liters: 25 mL / 1000 = 0.025 L)
0.83 M * 0.006 L = C2 * 0.025 L
0.00498 M·L = 0.025C2 M·L
Solving for C2: C2 = 0.00498 M·L / 0.025 L = 0.1992 M
The final concentration is 0.1992 M.
To calculate the final concentration of a solution, you can use the formula:
C1V1 = C2V2
Where:
C1 = Initial concentration of the solution
V1 = Initial volume of the solution
C2 = Final concentration of the solution
V2 = Final volume of the solution
Let's apply this formula to solve the problems:
a. 4.0 L of a 4.0 M HNO3 solution is added to water so that the final volume is 8.0 L.
Using the formula:
(4.0 M)(4.0 L) = C2(8.0 L)
Solving for C2:
C2 = (4.0 M)(4.0 L) / (8.0 L)
C2 = 2.0 M
Therefore, the final concentration of the solution is 2.0 M.
b. Water is added to 0.35 L of a 6.0 M KOH solution to make 2.0 L of a diluted KOH solution.
Using the formula:
(6.0 M)(0.35 L) = C2(2.0 L)
Solving for C2:
C2 = (6.0 M)(0.35 L) / (2.0 L)
C2 = 1.05 M
Therefore, the final concentration of the solution is 1.05 M.
c. A 20.0 mL sample of 8.0% (m/v) NaOH is diluted with water so that the final volume is 200.0 mL.
First, we need to calculate the mass of NaOH in the initial volume:
mass of NaOH = (8.0%)(20.0 mL) = 1.6 g
Then, we can use the formula to calculate the concentration in the final volume:
(1.6 g) / (200.0 mL) = C2
Converting mL to L:
(1.6 g) / (0.200 L) = C2
C2 = 8.0 g/L
Therefore, the final concentration of the solution is 8.0% (m/v) NaOH.
d. A 6.0 mL sample of 50.0% (m/v) acetic acid HC2H3O2 solution is added to water to give a final volume of 25 mL.
First, we need to calculate the mass of acetic acid in the initial volume:
mass of acetic acid = (50.0%)(6.0 mL) = 3.0 g
Then, we can use the formula to calculate the concentration in the final volume:
(3.0 g) / (25 mL) = C2
Converting mL to L:
(3.0 g) / (0.025 L) = C2
C2 = 120 g/L
Therefore, the final concentration of the solution is 120% (m/v) acetic acid HC2H3O2.
Note: The formula assumes that the volumes are additive and there is no volume change upon mixing.