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Preparing Solutions and Molarity

  1. Sep 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate how many grams of solute are necessary to prepare:
    1) 100 mL of 0.1 N (Normal Solution) Ca(OH)2 (molecular weight = 74)

    2) 100 mL 5% NaCl (molecular weight = 40)


    2. Relevant equations
    Molarity = # of moles / # of liters


    3. The attempt at a solution
    1. I set up dimensional analysis.

    74g Ca(OH)2 divided by 2 since there are 2 Hydrogens = 37g Ca(OH)2 for Molecular Weight

    100ml * (1 L / 1,000 mL) * (.05 mol / 1 L) * (37g / 1 mol) = 0.185g

    I'm not sure if it's really .05 mol/L for Ca(OH)2. I just figured since the molarity was 0.1 N that I could divide 0.1 by 2 hydrogens to give me .05 mol/L

    2. I honestly do not know how to even start this..
     
  2. jcsd
  3. Sep 8, 2012 #2

    AGNuke

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    Gold Member

    1.) Prepare 0.2 M solution of Calcium Hydroxide.

    2.) Just take 5 gms of NaCl, if we are talking about % w/v, which generally is the case.
     
  4. Sep 8, 2012 #3
    Why is it 0.2M for #1? And can you show me how to set up for #2 please? Thank you once again!
     
  5. Sep 9, 2012 #4

    Borek

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    Staff: Mentor

    Not 0.2, but 0.05 M Ca(OH)2. 0.05M solution will be 0.05M in Ca2+ and 0.1M in OH- - compare with the definition of normality.

    It is not clear if you are to prepare 5% w/w or w/v solution (not that the difference will be large). If w/v - by definition of the w/v it means 5 g in 100 mL of the solution. If w/w - approach depends on how accurate you want to be.

    Simple and less accurate approach - dilute solutions have density of 1 g/mL. That means you will need 100 g of solution. Of this 5% is NaCl and 95% is solvent (w/w). Use this information to calculate mass of NaCl.

    Exact approach (but most likely overkill) is to check the density of the solution in the density tables (for 5% NaCl it is 1.034 g/mL), calculate mass of the 100 mL - and do exactly the same thing you did in the above case.
     
  6. Sep 9, 2012 #5
    So for #1 would I set it up like this:

    # moles of Ca(OH)2 = 0.100 ml * {0.100 moles of OH(!)/l} * (1 mole of Ca(OH)2 / 2moles of OH)
     
  7. Sep 9, 2012 #6

    Borek

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    Staff: Mentor

    0.100 L, not 0.100 mL. Other than that, looks OK to me.
     
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