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Having a massive brain fart SIMPLE molarity -> molality conversion

  1. Nov 24, 2008 #1
    Having a massive brain fart... SIMPLE "molarity -> molality" conversion...

    This isn't even a homework question per se, I'm just trying to understand how the question converted the molarity to the molality value, via the density given. I just got a different value of molality when trying to solve for it manually... and am pretty sure I did something stupid.

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

    If the osmotic pressure of a 6.02×10-2 M aqueous solution of Ca(NO3)2 was found to be 4.05 atm at 20°C. Note that the corresponding molality would be 6.10×10-2 m, given that the density of the solution is 0.997 g/cm3....

    2. Relevant equations

    None.

    3. The attempt at a solution

    Okay.... so the question tells me it has a molality of 6.10×10-2 m,

    So if I want to solve for that myself, I would start off with the given molarity, right? So it would be 6.02×10-2 M then multiply it by the density of solution, 0.997g/cm3, or in other words, 0.997kg/1L ... and then we're done, right?

    So:
    6.02×10-2 mol/L multiply 1L/0.997kg = 6.04x10-2

    ... but the question tells me it's 6.10×10-2 m .... and I have no idea why I'm not getting that value.

    I don't know what's going wrong here... I'm sure I am missing something dead simple...
     
  2. jcsd
  3. Nov 25, 2008 #2

    Borek

    User Avatar

    Staff: Mentor

    Re: Having a massive brain fart... SIMPLE "molarity -> molality" conversion...

    Molality <> molarity conversion isn't that easy, proper formula is much more complicated.

    See example of concentration conversion - percentage to molarity. That's not exactly what you need, but the general approach outlined there will lead you to the molarity/molality formula. There is also a nice collection of conversion formulas on the same site.
     
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