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Homework Help: Molarity and Molality

  1. Nov 18, 2013 #1


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    Gold Member

    1. The problem statement, all variables and given/known data

    Screenshot of some prof's website. This is just ridiculous.

    http://i.minus.com/jnUa604Tz6bKX.png [Broken]

    2. Relevant equations

    Molarity to molality.

    Molality is moles / kg of solvent.

    Molarity is moles / volume of solution.

    3. The attempt at a solution

    Pretty sure all the ones in the screenshot from the webpage of some university professor are wrong.

    For 1)

    a) Assume 1000 mL of solution. We therefore have 0.84 moles of sucrose (0.84 moles / 1 L = 0.84 M)

    b) This means we have 1120 g of solution.

    c) Solution mass = solute mass + solvent mass.

    d) Solute mass = 0.84 moles * 342 g/moles = 287 grams.

    e) 1120 - 287 = solvent mass = 833 g = 0.833 kg.

    f) 0.84 / 0.833 isn't anywhere close to 3/4 molal solution.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 18, 2013 #2
    I see nothing wrong with your math, so I guess his answer is wrong.
  4. Nov 18, 2013 #3
    The website incorrectly assumed that molality is the number of moles of solute per kg of solution (rather than the correct definition which you gave).
  5. Nov 19, 2013 #4


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    Staff: Mentor

    Not to mention the fact density of 0.840 M sucrose solution is 1.1079 g/mL, density of 4.91 M NaOH solution is 1.1840 g/mL, and of 0.79 M NaHCO3 solution is 1.0455 g/mL. Neither is listed correctly (although sucrose is close).
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