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chemistry

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What volume ( L ) of the solute methanol should be mixed with 3.41 L of the solvent ethanol to make a solution that is 11.0 m methanol?

Am I supposed to use the molality eq'n?

  • chemistry -

    yes.
    m = mols solute/kg solvent
    You will need the densities to convert grams to volume and volume to grams.

  • chemistry -

    I converted ethanol using it's density, so I got mass=(.7893 kg/L)(3.41 L)=2.691513 g ethanol
    then plugged it into m=#mols solute/kg mass solvent to solve for # mols of solute which I got to be #mols=(11 m)(2.691513 g ethanol)=29.6066

    Now do I take this and multiply it with the molar mass of methanol, and then use the density relation to solve for volume?

  • chemistry -

    I wouldn't go that way. What you have done so far is very good. I would then determine how much methanol was needed.
    molality = moles CH3OH/0.269 kg solvent
    moles CH3OH = 11 x 0.269 = ??
    Then moles CH3OH x molar mass CH3OH = grams and convert that with density to volume CH3OH.

  • chemistry -

    Hmm okay, I followed that and is the answer that you got 119.795 L?

  • chemistry -

    Hang on a second. I've messed up in a couple of places. Let me get it straight an I will post back here.

  • chemistry -

    I have been using the wrong molar masses (I thought I knew them by memory but I didn't). Also, I used the wrong volume of ethanol. So here goes the correct way.

    3.41 L ethanol x 1000 = 3410 mL ethanol.
    m = v x d
    m = 3410 x 0.789 = 2960 grams or 2.96 kg ethanol.

    How much solute do we need?
    m = moles solute/kg solvent.
    11 = moles solute/2.69
    11 x 2.69 = 29.59 mols solute = 29.59 moles methanol.
    20.59 moles methanol x molar mass (32) = 946.9 grams methanol.
    m = v x d
    946.9 = v x 0.792 g/mL
    v = 1196 mL or 1.196 L.
    You may get slightly different values if you use slightly different molar masses and/or slightly different densities.

    Now let's check it to make sure we are right.
    1.196 L methanol/3.41 L ethanol.

    1.196 L methanol = 1196 mL
    1196 mL x 0.792 g/mL = 947 grams methanol.
    947/32 = 29.6 moles methanol.

    3.41 L ethanol solvent = 3410 grams.
    m = v*d
    m = 3410 x 0.789 g/mL = 2690 g ethanol or 2.69 kg ethanol.
    Then 29.6 mols solute/2.69 kg = 11.00 m.
    I'm glad you wrote that Hmmm--it made me go back and check the problem and my numbers. I hope I haven't confused you.

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