I've watched a tutor solve a simple dilution solutions problem on video.
There's a calculation in that problem I wish you could clarify for me.
It goes like this:

Approximately what volume of water is needed to dilute 25.0 mL of 18.0 M sulfuric acid to 3.0 M ?

M * L = M * L
(18.0 M)(25.0 mL) = (3.0 M)(V)
vol = 150 mL of solution.

I calculate the volume here to be 4.17 mL. It's a little tricky. Could you please explain.

Thank You.
Have a Great Night.

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asked by Cliff
  1. try finding the Moles/liter first. convert the 25.0mL to Moles/L. See what cha come up with.

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    posted by Denny
  2. I have an easier way to understand.

    you want 18M to 3M? that is a dilution of 6 times. So you want 1 part original, 5 parts water.

    so, what is one part: in this case it is given: 25ml
    so what is five parts: 125 ml.
    to give a new total volume of acid at 3M of 150ml.

    Try my method on these.

    What if you wanted to use 18M acid to make 500ml of acid at 2.2M?

    dilution factor 18/2.2 or 8.1818times
    one part original, 7.1818 water

    well, what is one part? 500ml/8.1818= 61.11ml.
    what is 7.1818 parts: 61.11*7.1818=(500 / 8.1818) * 7.1818 = 438.888753

    so, take 438.89 ml water, add 61.11 ml of the original acid (always add acid to water). and you have it.

  3. I don't know how you came up with 4.17 mL. That may be the tricky part.

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