How many mL of a 6 M HCl solution are required to prepare 200 mL of a 0.2 M solution?

v1=c2v2/c1

v1= (0.2M)(200ml)/(6M)
v1=6.67 ml HCl

To determine how many mL of a 6 M HCl solution are required to prepare 200 mL of a 0.2 M solution, we can use the formula:

C₁V₁ = C₂V₂

where C₁ and V₁ represent the concentration and volume of the initial solution (6 M HCl) and C₂ and V₂ represent the concentration and volume of the final solution (0.2 M solution).

Rearranging the formula to solve for V₁, we have:

V₁ = (C₂V₂) / C₁

Substituting the given values:

V₁ = (0.2 M × 200 mL) / 6 M

Simplifying:

V₁ = (0.2/6) × 200 mL

V₁ = 6.67 mL

Therefore, you would need approximately 6.67 mL of a 6 M HCl solution to prepare 200 mL of a 0.2 M solution.