how many milliliters of a stock solution of 2.00 M KNO3 would you need to prepare 100.0 mL of 0.150 M KNO3?
To determine how many milliliters of a stock solution of 2.00 M KNO3 are needed to prepare 100.0 mL of 0.150 M KNO3, we can use the formula:
M1V1 = M2V2
where:
M1 = molarity of the stock solution
V1 = volume of the stock solution
M2 = desired molarity of the final solution
V2 = desired volume of the final solution
Let's substitute the given values into the formula:
2.00 M * V1 = 0.150 M * 100.0 mL
Now we can solve for V1:
V1 = (0.150 M * 100.0 mL) / 2.00 M
V1 = 7.5 mL
Therefore, you would need 7.5 milliliters of the stock solution of 2.00 M KNO3 to prepare 100.0 mL of 0.150 M KNO3.
To find out how many milliliters of a stock solution of 2.00 M KNO3 you would need to prepare 100.0 mL of 0.150 M KNO3, you can use the concept of dilution.
The formula for dilution is as follows:
C1V1 = C2V2
Where:
C1 = initial concentration of the stock solution (2.00 M)
V1 = initial volume of the stock solution (unknown)
C2 = final concentration of the diluted solution (0.150 M)
V2 = final volume of the diluted solution (100.0 mL)
Now, let's substitute the given values into the formula:
(2.00 M)(V1) = (0.150 M)(100.0 mL)
Now, we can solve for V1, the initial volume of the stock solution:
V1 = (0.150 M)(100.0 mL) / (2.00 M)
V1 = 7.50 mL
So, you would need to measure out 7.50 mL of the stock solution with a concentration of 2.00 M in order to prepare 100.0 mL of a 0.150 M KNO3 solution.