How many milliliters of stock solution of 2.00 M KNO3 would you need to prepare 100.0 ml of .20 M KNO3 solution? Please show work/explain the answer Thank You!
To determine the volume of the stock solution needed to prepare a given volume of a diluted solution, we can use the dilution equation:
C1V1 = C2V2
where C1 and V1 are the concentration and volume of the stock solution, and C2 and V2 are the concentration and volume of the diluted solution.
In this case:
C1 = 2.00 M (concentration of stock solution)
V1 = unknown
C2 = 0.20 M (concentration of diluted solution)
V2 = 100.0 ml (volume of diluted solution)
Let's plug in the values into the equation and solve for V1:
C1 * V1 = C2 * V2
2.00 M * V1 = 0.20 M * 100.0 ml
Now, we can solve for V1:
V1 = (0.20 M * 100.0 ml) / 2.00 M
V1 = 10.0 ml
Therefore, you would need 10.0 ml of the 2.00 M stock solution to prepare 100.0 ml of a 0.20 M KNO3 solution.
To find out how many milliliters of stock solution of 2.00 M KNO3 are needed to prepare a 100.0 ml of 0.20 M KNO3 solution, we can use the formula:
C₁V₁ = C₂V₂
Where:
C₁ is the concentration of the stock solution (2.00 M)
V₁ is the volume of the stock solution we need to find
C₂ is the desired concentration of the diluted solution (0.20 M)
V₂ is the desired final volume of the diluted solution (100.0 ml)
We can rearrange the formula to solve for V₁:
V₁ = (C₂ * V₂) / C₁
Substituting the given values:
V₁ = (0.20 M * 100.0 ml) / 2.00 M
V₁ = (20.0 ml) / 2.00
V₁ = 10.0 ml
Therefore, you would need 10.0 ml of the stock solution of 2.00 M KNO3 to prepare 100.0 ml of 0.20 M KNO3 solution.
The easiest and quickest way to solve these dilution problems is to use the dilution formula.
mL1 x M1 = mL2 x M2
mL1 x 2.00M = 100.0 mL x 0.20 M
Solve for mL 1