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Math Help - dilute concentration

  1. #1
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    Question dilute concentration

    Hello, I have a question regarding dilution:

    If i need to dilute 50% dextrose into 20mg of a 2.5% solution, how would i do that?
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  2. #2
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    If i need to dilute 50% dextrose into 20mg of a 2.5% solution, how would i do that?

    First, convert your %solution into a Molarity: [(%solution)*10]/FW

    This answer gets you g/L

    Since Dextrose (or d-glucose) is C6H12O6, it has a formula weight (FW) of 180.16 using your periodic table.

    \frac{(50g/100ml)*10}{180.16 g/L} = 0.02775M (keep the sig digits for now)

    Do the same for a 2.5% solution: you get 0.00139M

    Finally, convert 20mg to of 0.00139M C6H12O6 into liters, and use your friend M1V1 = M2V2 (M is molarity and V is volume).

    I am three years removed from all this, so double check your and my work. I am assuming this "explanation" is better nothing. Especially check units, and that is always.
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  3. #3
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    Hello, starcherub08!

    If i need to dilute 50% dextrose into 20mg of a 2.5% solution,
    how would i do that?
    Since we will be adding pure water, consider the amount of water at each stage.

    Let x = amount of water to be added.
    Then 20-x = amount of the 50% solution.

    We start with 20-x mgs of solution which is 50% water.
    . . It contains: . 0.5(20-x) mgs of water.

    We add x mgs of pure water.
    . . It contains (of course): . x mgs of water.

    The mixture contains: . \boxed{0.5(20-x) + x} mgs of water. .[1]


    Consider the mixture again.
    . . It will be 20 mgs of solution which is 97.5% water.
    The mixture contains: . 0.975(20)\:=\:\boxed{19.5} mgs of water. .[2]


    We just described the final amount of water in two ways.

    There is our equation: [1] = [2] . . . . {\bf0.5(20-x) + x \:=\:19.5}

    . . Got it?

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