In the laboratory you dilute 5.09 mL of a concentrated 12.0 M perchloric acid solution to a total volume of 150 mL. What is the concentration of the dilute solution?

Answer 1

Formula of perchloric acid is HClO4,

If in 1 L of acid solution 12 mol of HClO4 then in 5.09 mL or 0.0051 L is 0.06 mol of acid

If we dilute the solution or add water, we will have 150 mL or 0.150 L of solution, where still dissolved 0.06 mol of acid.

now we can find a new changed concentration because we added water

0.06 mol/ 0.150L =0.4 M

Answer 2

It asks you to find the concentration, which is safe to assume that the unit is Molarity (M). Also, this unit was used in a the problem as well.

Molarity has units of moles/L so we have to work on getting to this point.

The problem states that you have 5.09 mL of a 12.0 M concentration. First, we will need moles. To get moles, we use stoichometry:

5.09 mL = 0.00509 L

0.00509 L x ( 12 moles/ L) = 0.06108 moles

This is how many moles are present in the solution. To find the new concentration, we divide the total number of moles by the new volume in Liters:

0.06108 moles/0.150L = 0.4072 M

Rounding to two sig.figs, we get 0.41 M.

Answer 3

Simple method is to use V1M1 = V2M2

V1 = initial volume = 5.09 ml

M1 = 12.0 M

V2 = 150 ml

M2 = ?

(5.09)(12.0) = (150 ml)(x M)

x = 0.407 M = 0.41 M (to 2 significant figures based on 2 s.f. in 150 ml)