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it is desired to prepare 250.0mL of a standard solution having a concentration of 0.1000M AgNo3. How many grams of a sample of AgNO3 of 98.8% purity are required for this purpose?

  • chemistry -

    M = moles/L
    0.1 = moles/0.250 L
    solve for moles.

    moles = grams/molar mass
    You know moles and molar mass, solve for grams (100% pure) AgNO3.

    grams impure AgNO3 x 0.988 = grams pure AgNO3 you want or rearrange to
    grams pure AgNO3/0.988 = grams you must weigh out to get grams AgNO3 you need.

  • chemistry -

    the answer in the book was 4.3 grams
    but if you do that you get
    169.8731/0.988 which is 171.94 g
    i don't get it

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