What is the volume, in milliliters, required to prepare
a. 20.0 mL of a 0.250 M KNO3 solution using a 6.00 M KNO3 solution
b. 25.0 mL of a 2.50 M H2SO4 solution using a 12.0 M H2SO4 solution
c. 0.500 L of a 1.50 M NH4Cl solution using a 10.00 M NH4Cl solution
a. Well, it's all about mixing and matching! To prepare 20.0 mL of a 0.250 M KNO3 solution, you could take 1 part of the 6.00 M KNO3 solution and dilute it with 23 parts of water. Why 23? Because chemistry always likes to keep things interesting, I suppose.
b. Ah, the art of potion-making! To make 25.0 mL of a 2.50 M H2SO4 solution, you'll need to mix 2 parts of the 12.0 M H2SO4 solution with 3 parts of water. It's like creating the perfect balance between acidity and hydration. Just don't forget to stir it with a magic wand!
c. Ah, the mysterious world of solutions! To create 0.500 L of a 1.50 M NH4Cl solution, you'll need to combine 0.075 L of the 10.00 M NH4Cl solution with 0.425 L of water. It's like mixing an intense ammonia smell with a refreshing rain shower. Voilà! You've got yourself a solution.
To calculate the volume of a solution needed to prepare a desired concentration, we can use the formula:
(C1)(V1) = (C2)(V2)
Where:
C1 = initial concentration (given)
V1 = initial volume (unknown)
C2 = final concentration (given)
V2 = final volume (desired)
a. To prepare 20.0 mL of a 0.250 M KNO3 solution using a 6.00 M KNO3 solution:
Using the formula, we have:
(6.00 M)(V1) = (0.250 M)(20.0 mL)
Rearranging the formula to solve for V1, we get:
V1 = (0.250 M)(20.0 mL) / 6.00 M
V1 ≈ 0.833 mL
Therefore, you would need approximately 0.833 mL of the 6.00 M KNO3 solution to prepare 20.0 mL of a 0.250 M KNO3 solution.
b. To prepare 25.0 mL of a 2.50 M H2SO4 solution using a 12.0 M H2SO4 solution:
Using the formula, we have:
(12.0 M)(V1) = (2.50 M)(25.0 mL)
Rearranging the formula to solve for V1, we get:
V1 = (2.50 M)(25.0 mL) / 12.0 M
V1 ≈ 5.21 mL
Therefore, you would need approximately 5.21 mL of the 12.0 M H2SO4 solution to prepare 25.0 mL of a 2.50 M H2SO4 solution.
c. To prepare 0.500 L of a 1.50 M NH4Cl solution using a 10.00 M NH4Cl solution:
First, convert the desired volume to milliliters (mL):
0.500 L = 500 mL
Using the formula, we have:
(10.00 M)(V1) = (1.50 M)(500 mL)
Rearranging the formula to solve for V1, we get:
V1 = (1.50 M)(500 mL) / 10.00 M
V1 = 75 mL
Therefore, you would need 75 mL of the 10.00 M NH4Cl solution to prepare 0.500 L (500 mL) of a 1.50 M NH4Cl solution.
To determine the volume of a concentrated solution required to prepare a desired volume and concentration of a solution, you can use the formula:
V1 * C1 = V2 * C2
where:
V1 = volume of concentrated solution
C1 = concentration of concentrated solution
V2 = desired volume of final solution
C2 = desired concentration of final solution
Now, let's calculate the required volumes for each scenario:
a. To prepare 20.0 mL of a 0.250 M KNO3 solution using a 6.00 M KNO3 solution:
V1 * 6.00 M = 20.0 mL * 0.250 M
Rearranging the formula to solve for V1:
V1 = (20.0 mL * 0.250 M) / 6.00 M
V1 ≈ 0.833 mL
Therefore, you need approximately 0.833 mL of the 6.00 M KNO3 solution.
b. To prepare 25.0 mL of a 2.50 M H2SO4 solution using a 12.0 M H2SO4 solution:
V1 * 12.0 M = 25.0 mL * 2.50 M
Rearranging the formula to solve for V1:
V1 = (25.0 mL * 2.50 M) / 12.0 M
V1 ≈ 5.21 mL
Therefore, you need approximately 5.21 mL of the 12.0 M H2SO4 solution.
c. To prepare 0.500 L of a 1.50 M NH4Cl solution using a 10.00 M NH4Cl solution:
V1 * 10.00 M = 0.500 L * 1.50 M
Rearranging the formula to solve for V1:
V1 = (0.500 L * 1.50 M) / 10.00 M
V1 = 0.075 L
Since the question asks for the volume in milliliters, we need to convert liters to milliliters:
V1 = 0.075 L * 1000 mL/L
V1 = 75 mL
Therefore, you need exactly 75 mL of the 10.00 M NH4Cl solution.
20.0 ml of a 0.250 solution M KNO soul using a 6.00 M KNO solution.Determine the volum, in milliliters, required to prepare each of the following.
(a) you are diluting the concentration by a factor of 6/.25=24
So, you will need 20/24 ml of 6M solution to produce 20ml of .25M solution. The rest, of course, is water.
Do the others in like wise.