How much of a 500 ml bottle of 1.00 molar KCI solution should be diluted with deionized water to give 200.0 ml of a 0.100 molar solution

To solve this problem, we need to use the concept of dilution in chemistry.

Dilution refers to the process of reducing the concentration of a solute in a solution. It involves adding more solvent (in this case, deionized water) to an existing solution to decrease the concentration of the solute.

In this case, we want to know how much of a 500 mL bottle of a 1.00 molar KCI (potassium chloride) solution should be diluted with deionized water to obtain 200.0 mL of a 0.100 molar solution.

First, let's define the formula to calculate dilutions:
C1V1 = C2V2

Where:
C1 = initial concentration of the solution (in this case, 1.00 molar)
V1 = volume of the solution that we will take out (unknown)
C2 = final concentration of the solution (0.100 molar)
V2 = final volume of the solution (200.0 mL = 0.200 L)

Using this formula, we can rearrange it to solve for V1:
V1 = (C2V2) / C1

Now, we can plug in the known values:
V1 = (0.100 mol/L * 0.200 L) / 1.00 mol/L

V1 = 0.020 liters = 20 mL

Therefore, you need to take out 20 mL of the 500 mL bottle of the 1.00 molar KCI solution and dilute it with deionized water to obtain 200.0 mL of a 0.100 molar solution.