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4 February, 10:06

If you dilute 13.0 ml of the stock solution to a final volume of 0.260 l, what will be the concentration of the diluted solution?

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  1. 4 February, 10:13
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    V = 0.260 L = 260 mL

    If you give the initial concentration a number, any number, then use C = n/V or n = C*V, one thing you will notice is that n does not change. So your formula becomes

    C*V = Cf * Vf

    C = 2 mol/L (this number is just made up.)

    V = 13 mL

    C1 = ? (you have not added anything but water).

    V1 = 260 mL

    2 * 13 = 260 x

    26 = 260 x

    x = 0.100 mol/L

    the new concentration is 0.1 moL/L The ratio of old to new is 2/0.1 = 20 which means that the old concentration is 20 times the new.

    Now we have to do it the more expected way. n is still the same during the dilution.

    C = C

    V = 13 mL

    C1 = ?

    V1 = 260 mL

    C * 13 = C1 * 260 mL

    C / C1 = 260 / 13

    C / C1 = 20

    The old concentration was 20 times the diluted one. If you are having trouble with this idea, drop me a note. I'll see what will get you to understand it.
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