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.