How much liquid is needed to prepare 629.1mL of a solution that has a new concentration of 11.2M if the stock solution is 26.1M

270. mL

General Formulas and Concepts:

Acid-Base Titrations

Dilution: M₁V₁ = M₂V₂

M₁ is stock molarityV₁ is stock volumeM₂ is new molarityV₂ is new volume

Explanation:

Step 1: Define

Identify

[Given] V₂ = 629.1 mL

[Given] M₂ = 11.2 M

[Given] M₁ = 26.1 M

[Solve] V₁

Step 2: Find Stock Volume

Substitute in variables [Dilution]:                                                                     (26.1 M)V₁ = (11.2 M)(629.1 mL)Multiply:                                                                                                             (26.1 M)V₁ = 7045.92 M · mLIsolate V₁ [Cancel out units]:                                                                            V₁ = 269.959 mL

Step 3: Check

Follow sig fig rules and round. We are given 3 sig figs as our lowest.

269.959 mL ≈ 270. mL

Alpha particles have positive charge, beta particles have negative charge, and answer 4 just doesn’t make sense. thus, you are left with only one possible answer.