Question

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

Answer 1

37.8 mL

Step-by-step explanation:

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)

You 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.