i need 200mL of an 8 molar solution of dilute acid, and my stock cconcentrate is 18 molar, how much of the concentrate, and how much water will i need?

To prepare the desired 8 molar solution of dilute acid, you need to determine the amount of concentrate acid and water required. You can use the concept of dilution to calculate this.

The formula for dilution is:

C1V1 = C2V2

Where:
C1 = Initial concentration of the stock solution
V1 = Volume of the stock solution used
C2 = Final concentration of the diluted solution
V2 = Final volume of the diluted solution

Let's apply this formula to your case:

Given:
C1 = 18 M (concentration of stock solution)
C2 = 8 M (desired concentration of diluted solution)
V2 = 200 mL (final volume of diluted solution)

First, we need to find V1, the volume of the stock solution required to make the desired dilution.

Rearranging the formula, we have:

V1 = (C2V2) / C1

Substituting the values:

V1 = (8 M * 200 mL) / 18 M
V1 = 88.89 mL

So, you will need 88.89 mL of the 18 M stock solution.

To determine the amount of water needed, subtract the volume of the concentrate from the final volume of the diluted solution:

Water volume = V2 - V1
Water volume = 200 mL - 88.89 mL
Water volume ≈ 111.11 mL

Therefore, to prepare 200 mL of an 8 molar solution of dilute acid with an 18 molar stock concentrate, you will need approximately 88.89 mL of the concentrate and about 111.11 mL of water.