What volume of a 6.0 M NaOH solution would you need to prepare 250.0 milliliters of a 3.0 M NaOH solution?
How many moles do you need? That is M x L = 3.0M x 0.250 L= ??
Now M NaOH = moles NaOH/L NaOH and solve for L NaOH, then convert to mL.
An easier way is to use the dilution formula
c1v1 = c2v2
6*x = 3.0*250
solve for x
To calculate the volume of a 6.0 M NaOH solution needed to prepare 250.0 milliliters of a 3.0 M NaOH solution, we can use the formula:
M₁V₁ = M₂V₂
where M₁ and V₁ are the initial concentration and volume, and M₂ and V₂ are the final concentration and volume.
Let's substitute the given values:
M₁ = 6.0 M (initial concentration)
V₁ = ? (initial volume)
M₂ = 3.0 M (final concentration)
V₂ = 250.0 mL (final volume)
Now we can solve for V₁:
(6.0 M)(V₁) = (3.0 M)(250.0 mL)
Divide both sides by 6.0 M to isolate V₁:
V₁ = (3.0 M)(250.0 mL) / 6.0 M
V₁ ≈ 125.0 mL
Therefore, you would need to prepare approximately 125.0 milliliters of a 6.0 M NaOH solution to obtain 250.0 milliliters of a 3.0 M NaOH solution.