How many mL of a 2.00 M NaF solution are required to make 450 mL of a 0.168 M NaF solution?

Let x be the amount of the 2.00 M NaF solution needed.

Using the formula for dilution:
M1V1 = M2V2
where
M1 = 2.00 M (concentration of the stock solution)
V1 = x mL (volume of the stock solution added)
M2 = 0.168 M (concentration of the final solution)
V2 = 450 mL (volume of the final solution)
We can solve for x:
2.00 M * x mL = 0.168 M * 450 mL
x = (0.168 M * 450 mL) / 2.00 M
x = 37.8 mL
Therefore, 37.8 mL of the 2.00 M NaF solution are required to make 450 mL of a 0.168 M NaF solution.