How much concentrated 12 M HCl is needed to prepare 200 ml of a 2.5 M solution?

you are diluting it 12/2.5 times, or 4.8 times, which means you add to one part of concentrate acid, 3.8 part water.

so what is one part? 200m/4.8= 41.67 ml.

So add 41.67ml of the 12 M HCl to 150 ml water, while steering wearing googles, then when cool, and the rest of the water up to 200ml of solution. put it in a suitable decanter, label it properly.

SNOOT

To determine the amount of concentrated 12 M HCl needed to prepare a 2.5 M solution, we can use the formula:

C1V1 = C2V2

Where:
C1 = concentration of the concentrated HCl solution
V1 = volume of the concentrated HCl solution needed
C2 = desired concentration of the final solution
V2 = volume of the final solution

Plugging in the given values:
C1 = 12 M
V1 = ?
C2 = 2.5 M
V2 = 200 ml

Now, let's solve for V1:

C1V1 = C2V2
12 M(V1) = 2.5 M(200 ml)

Divide both sides by 12 M:
V1 = (2.5 M * 200 ml) / 12 M

V1 ≈ 41.67 ml

Therefore, approximately 41.67 ml of the concentrated 12 M HCl is needed to prepare 200 ml of a 2.5 M solution.

To determine the amount of concentrated 12 M HCl needed to prepare a 2.5 M solution, we can use the equation:

M1 × V1 = M2 × V2

Where:
M1 = initial concentration of the solution (12 M HCl)
V1 = initial volume of the solution (unknown)
M2 = final concentration of the solution (2.5 M HCl)
V2 = final volume of the solution (200 ml)

We can rearrange the equation to solve for V1 as follows:

V1 = (M2 × V2) / M1

Substituting the values into the equation:

V1 = (2.5 M × 200 ml) / 12 M

V1 = (500 ml⋅M) / 12 M

Simplifying the units:

V1 ≈ 41.67 ml

Therefore, approximately 41.67 ml of 12 M HCl is needed to prepare 200 ml of a 2.5 M solution.