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Concentration problem

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the required mass of Al2(SO4)3.18H2O (molar mass = 666 g/mol) to prepare 50 mL of an aqueous solution, concentration = 40 mg of Al3+ per mL


    3. The attempt at a solution
    0.04 grams of Al3+/1mL = x g / 50 mL
    x = 2g

    I think this is the required mass of Al3+, now I don't know how to find how much Al2(SO4)3.18H2O do I need to get 2 g of Al3+

    The answer is 24.7g
     
  2. jcsd
  3. Feb 12, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    How many moles of Al do you need?

    In how many moles of aluminum sulfate there will be required amount of Al?

    What will be mass of these moles?
     
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