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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

Sagot :

Spaceidn

Answer:

270. mL

General Formulas and Concepts:

Acid-Base Titrations

Dilution: M₁V₁ = M₂V₂

  • M₁ is stock molarity
  • V₁ is stock volume
  • M₂ is new molarity
  • V₂ 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

  1. Substitute in variables [Dilution]:                                                                     (26.1 M)V₁ = (11.2 M)(629.1 mL)
  2. Multiply:                                                                                                             (26.1 M)V₁ = 7045.92 M · mL
  3. Isolate 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

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