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1. How many grams of CaCl2 would be required to produce a 3.5 M (molar) solution with a volume of 2.0 L?

  • Chemistry -

    How many mols do you want? That's mols = M x L = ?
    Then mol = grams/molar mass. You know molar mass and mols, solve for grams.

  • Chemistry -


    Is that right? /:

  • Chemistry -

    The first part is right. The second is not.
    Yes, M x L = mols = 3.5 x 2L = 7 mols

    Then mol = g/molar mass
    7 = grams/molar mass CaCl2.
    grams = 7*molar mass CaCl2 = estimated 7*111 = about 777 grams.

  • Chemistry -

    I still don't get it. How or where'd you get 111 from? /:

  • Chemistry -

    Dr. Bob222 gave you the setup to calculate the grams of CaCl2.

    But I will go into a little bit more detail and maybe this will help:

    Molarity=Moles/Volume (L)

    The question tells you that you need need a total of 2L, but you don't know how many moles are in a 3.5M solution with a volume of 2L, but you can find that out: how do you calculate the moles needed to make a 3.5M solution that has a volume of 2L?

    3.5M=x moles/2L

    Solve for X:

    3.5*2L=x moles

    ***You did this correctly, but you have to convert moles to grams because the question asks you for grams.

    Stoichiometry allows you to do this, so how do you convert moles to grams?

    x moles * molecular weight

    x moles * 111g of CaCl2/mole=???????

    I'll let you take care of it from here.

  • Chemistry -

    111 is the approximate molar mass of CaCl2.
    Ca = about 40 from the periodic table.
    Cl = about 35.5 and 2*35.5 = about 71
    71+40 = about 111. You can look up the numbers and obtain a more accurate answer. You should be able to look at the math equation I wrote above (mols = g/molar mass CaCl2 and g = mols x molar mass CaCl2) to see that I substituted 7 for mols from the first part and that I substituted 111 for molar mass. Then molar mass x mols = 111 x 7 = about 777. What I'm saying is that you should be able to tell that the 111 MUST be for the molar mass of CaCl2.

  • Chemistry -

    Oh.Oh okay. I get it. Thanks!

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