How much water should be added to 125.16 mL of 18.0 M H2SO4 to prepare a 1.27 M solution?

Answer 1

For this type of problem, you can use V1M1 = V2M2

V1 = initial volume = 125.16 ml

M1 = initial molarity = 18.0 M

V2 = final volume = ?

M2 = final molarity = 1.27 M

Solving for V2, we have...

(125.16)(18.0) = (V2)(1.27)

V2 = 1774 mls .. NOTE: this is the FINAL volume

Since we began with a volume of 125.16 mls, the volume that must be added is 1774 mls - 125 mls

= 1649 mls = 1650 mls (3 sig. figs.) must be added

Answer 2

We can use the formula V1M1 = V2M2, where:

V1 = initial volume = 125.16 mL (3 sig figs)

M1 = initial molarity = 18.0 M (3 sig figs)

V2 = final volume

M2 = final molarity = 1.27 M (3 sig figs)

We want to find the final volume (V2) and then calculate the volume of water to be added.

Solving for V2, we have:

(125.16)(18.0) = (V2)(1.27)

V2 = (125.16 * 18.0) / 1.27

V2 ≈ 1775 mL (3 sig figs, since the limiting factor is the final molarity with 3 sig figs)

Since we began with a volume of 125.16 mL, the volume that must be added is:

1775 mL - 125.16 mL ≈ 1650 mL (3 sig figs)

So, 1650 mL of water must be added to the 125.16 mL of 18.0 M H2SO4 to prepare a 1.27 M solution, considering 3 significant figures.