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Solutions and concentrationChemistryHigh schoolabout 0:52

How much 5 M stock solution do you need to make 400 mL of 1.25 M solution?: animated lesson

A narrated, step-by-step solutions and concentration video lesson with captions. Final answer: initial volume V₁ = 100 mL; add 300 mL of solvent.

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.3 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Worked example
    How much 5 M stock solution do you need to make 400 mL of 1.25 M solution?
  2. 2. Step by step
    1. M1V1=M2V2M_1V_1 = M_2V_2
      Moles of solute stay the same when you dilute
    2. V1=M2 V2M1V_1 = \frac{M_2\,V_2}{M_1}
      Rearrange for initial volume V₁
    3. V1=(1.25 M)(400 mL)5 M=100 mLV_1 = \frac{(1.25\ \text{M})(400\ \text{mL})}{5\ \text{M}} = 100\ \text{mL}
      Substitute
    4. V2−V1=300 mLV_2 - V_1 = 300\ \text{mL}
      Solvent to add
  3. 3. The answer
    initial volume V₁ = 100 mL; add 300 mL of solvent\text{initial volume V₁ = 100 mL; add 300 mL of solvent}

    initial volume V₁ = 100 mL

  4. 4. Check
    • M₁V₁ = 500 and M₂V₂ = 500 ✓

Transcript

The problem. Here is the problem: How much 5 M stock solution do you need to make 400 mL of 1.25 M solution? Diluting adds solvent but not solute, so the moles of solute before and after are equal.

Step by step. Moles of solute stay the same when you dilute. Rearrange for initial volume V₁. Substitute. Solvent to add. Start with V₁ of the stock and add solvent up to V₂.

The answer. So the answer is initial volume V₁ equals 100 mL; add 300 mL of solvent. Any volume unit works in M₁V₁ equals M₂V₂ as long as both volumes use the same one. Moles of solute: 0.5 mol.

Check. M₁V₁ equals 500 and M₂V₂ equals 500, which checks out.

About solutions and concentration

Molarity, dilution and moles of solute in a solution.

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