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Math Help - "Chemistry" algebra

  1. #1
    Member WhoCares357's Avatar
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    "Chemistry" algebra

    I have this problem for homework:
    How many mL of 1M acetic acid (C2H4O2) and .2 molal acetic acid need to be added to make 100 mL of vinegar (5% acetic acid)? Assume 1g/mL density of all solutions.
    It's a chemistry problem, but it's really just basic algebra. The different units represent the different variables.

    I started by listing everything that is given: density = 1g/mL; 1M solute = 1 mol solute/1 L solution; .2 molal solute = .2mol solute/1kg solvent (H2O); 60.052 g/mol solute; 18.000 g/mol solvent; .1L total solution. I am looking for the mL of solute in .1L of solution.

    My first step was to use the Molarity to figure out the number of moles of the solute: 1M solute = .1 mol solute/.1L solution.
    Then I used the number of moles to find the number of grams in the solution: .1mol x 60.052g/mol = 6.0052g solute.
    Finally I converted the mass to volume using density: 6.0052g/(1g/mL) = 6.0052mL.

    However I am pretty sure that this answer is wrong. When I insert the .1 mol of solute into the molality formula I get a totally different answer.
    .2 molal = .1mol/500g -> 500gx.05/.95 = 26.315g -> 26.315g/(1g/mL) = 26.315mL.

    26.315mL=/= 6.0052mL


    Can anyone show me where I went wrong. It's possible that I may have just overthought it, but the answer just doesn't seem right to me.

    Thank you.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by WhoCares357 View Post
    I have this problem for homework: It's a chemistry problem, but it's really just basic algebra. The different units represent the different variables.

    I started by listing everything that is given: density = 1g/mL; 1M solute = 1 mol solute/1 L solution; .2 molal solute = .2mol solute/1kg solvent (H2O); 60.052 g/mol solute; 18.000 g/mol solvent; .1L total solution. I am looking for the mL of solute in .1L of solution.

    My first step was to use the Molarity to figure out the number of moles of the solute: 1M solute = .1 mol solute/.1L solution.
    Then I used the number of moles to find the number of grams in the solution: .1mol x 60.052g/mol = 6.0052g solute.
    Finally I converted the mass to volume using density: 6.0052g/(1g/mL) = 6.0052mL.

    However I am pretty sure that this answer is wrong. When I insert the .1 mol of solute into the molality formula I get a totally different answer.
    .2 molal = .1mol/500g -> 500gx.05/.95 = 26.315g -> 26.315g/(1g/mL) = 26.315mL.

    26.315mL=/= 6.0052mL


    Can anyone show me where I went wrong. It's possible that I may have just overthought it, but the answer just doesn't seem right to me.

    Thank you.
    The concentration of molar acetic acid is ~=6%, and of 0.2 molar is ~=1.2% (by mass)

    How much of each must be mixed to give 100mL of 5% solution?

    CB
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by WhoCares357 View Post
    I have this problem for homework: It's a chemistry problem, but it's really just basic algebra. The different units represent the different variables.

    I started by listing everything that is given: density = 1g/mL; 1M solute = 1 mol solute/1 L solution; .2 molal solute = .2mol solute/1kg solvent (H2O); 60.052 g/mol solute; 18.000 g/mol solvent; .1L total solution. I am looking for the mL of solute in .1L of solution.

    My first step was to use the Molarity to figure out the number of moles of the solute: 1M solute = .1 mol solute/.1L solution.
    Then I used the number of moles to find the number of grams in the solution: .1mol x 60.052g/mol = 6.0052g solute.
    Finally I converted the mass to volume using density: 6.0052g/(1g/mL) = 6.0052mL.

    However I am pretty sure that this answer is wrong. When I insert the .1 mol of solute into the molality formula I get a totally different answer.
    .2 molal = .1mol/500g -> 500gx.05/.95 = 26.315g -> 26.315g/(1g/mL) = 26.315mL.

    26.315mL=/= 6.0052mL


    Can anyone show me where I went wrong. It's possible that I may have just overthought it, but the answer just doesn't seem right to me.

    Thank you.
    The concentration of molar acetic acid is ~=6%, and of 0.2 molar is ~=1.2% (by mass)

    How much of each must be mixed to give 100mL of 5% solution?

    CB
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  4. #4
    Member WhoCares357's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    The concentration of molar acetic acid is ~=6%, and of 0.2 molar is ~=1.2% (by mass)

    How much of each must be mixed to give 100mL of 5% solution?

    CB
    Edit: I think I figured out how you got the concentrations. I used the given volume for both formulas (molarity and molality) and got ~6g and ~1.14g.

    I don't understand what I have to do with them now, though. Do I just add them?
    Last edited by WhoCares357; August 29th 2010 at 07:09 AM.
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