in order to prepare 50.0ml of .200m hcl you will add ____ ml of 1.00m hcl to ___ml of water

How would I solve this equation?

Thanks!

Yes but read further. I assume the m is M and not m. M is molarity. m is molality. Use the dilution formula.

mL1 x M1 = mL2 x M2
? mL x 1.0 M = 50 mL x 0.200 M
? mL = (50 x 0.2/1) = 10 mL
Technically, one adds the 10 mL of the 1.0 M solution to a 50 mL volumetric flask, then add distilled water to the mark on the volumetric flask. That's the way you do it right. BUT, problems like this USUALLY say to assume the volumes are additive in which case you would add 40 mL H2O.

Would I add 10.0ml of 1.0m HCl which would be to 40ml of water?

To solve this equation, you need to use the formula for calculating the molarity of a solution:

M₁V₁ = M₂V₂

Where:
M₁ = molarity of the stock solution (1.00 M HCl)
V₁ = volume of the stock solution to be used (unknown)
M₂ = desired molarity of the solution (0.200 M HCl)
V₂ = volume of the final solution (50.0 mL)

Rearranging the formula, we get:

V₁ = (M₂ * V₂) / M₁

Now, substitute the given values into the formula:

V₁ = (0.200 M * 50.0 mL) / 1.00 M

V₁ = (0.200 * 50.0) / 1.00

V₁ = 10.0 mL

Therefore, to prepare 50.0 mL of a 0.200 M HCl solution, you will add 10.0 mL of 1.00 M HCl to 40.0 mL of water.

you stupid