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I've been working on this problem for hours and cant figure it out. I know it should be simple, but I'm stuck...

The question is...

Describe how to prepare 1.00L of a 1.0% (w/v) NaOH from a 2.0 M NaOH. *I need to figure out how many ml of the stock solution is needed.

So far I know
1. 1.0% (w/v) means 1g NaOH per 100 ml.

2. 2 M NaOH = 79.994 g NaOH

  • Chemistry -

    I would convert 1.0% w/v to M.
    1 g/100 mL = 10 g/1000 mL.
    10 g = ? mols; that's 10/40 = 0.25 mols and that in 1 L = 0.25M, then apply the dilution formula of
    c1v1 = c2v2
    0.25*1000mL = 2.0*v2mL
    solve for v2.

  • Chemistry -


    Where did you get 10/40? Also, where would the 79.994g of NaOH come in?

  • Chemistry -

    mols NaOH = grams NaOH/molar mass NaOH.
    Sorry I didn't spell that out.

  • Chemistry -

    Here is what I came up with, but not sure if it is correct.

    2 moles of NaOH = 79.994g NaOH

    79.994g/1L * 1L/1000mL * 100mL/1g = 7.9

  • Chemistry -

    I worked the problem for you except for the last step. All you need to do is to solve for v2 and the answer will be in mL. You're to describe how to make it You take v2 mL of the stock solution, place in a 1.0 L volumetric flask, then add water to the mark and mix thoroughly.

  • Chemistry -

    Okay. Thanks for your help.

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