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Solubility homework question

  1. Jul 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Which of the following aqueous solutions has a Cl– concentration of 2 m at 25°C?
    A.

    271 g of HgCl2 in 55 mol of H2O
    B.

    271 g of HgCl2 in 22.5 mol of H2O
    C.

    53 g of NH4Cl in 55 mol of H2O
    D.

    53 g of NH4Cl in 22.5 mol of H2O.

    Correct Answer
    Explanation:
    D. Pure water has a concentration of 55 M and density of 1 kg/L. 55 moles of water represent 1 L (1 kg) and 22.5 moles of water represent 0.5 L (0.5 kg). HgCl2 demonstrates poor water solubility and will dissociate little. Therefore, 1 mol of HgCl2 placed in either water 1 kg or 0.5 kg of water will not achieve a 2 m Cl– solution, eliminating choices A and B. NH4Cl is water soluble and has a molecular weight of 53 g/mol. Choice C is a 1 m NH4Cl (1 m Cl–) solution and Choice D is a 2 m NH4Cl (2 m Cl–) solution.


    2. Relevant equations

    No equation. Conceptual

    3. The attempt at a solution

    I understand that we are looking for the answer that has a concentration of 2 M for chlorine and I understand that A and B are wrong because that compound is insoluble in water but what I don't understand is why we get a 2 molar solution when we add it to half a liter of water vs getting a 1 molar solution when we add it to 1 liter of water
     
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  3. Jul 4, 2013 #2

    Borek

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    Staff: Mentor

    What is the definition of molar concentration?

    Have you tried to calculate concentrations for C and D?
     
  4. Jul 4, 2013 #3
    Oh yes I understand now. I got confused because I thought we had to factor in temperature somehow. But yes I see how you get a two molar concentration and a 1 molar concentration when you solve for moles and liters and solve for molarity. Thanks.
     
  5. Jul 4, 2013 #4

    chemisttree

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    Science Advisor
    Homework Helper
    Gold Member

    Is that molar or molal?
     
  6. Jul 5, 2013 #5

    Borek

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    Staff: Mentor

    Good point.

    Not that it changes much, as the values used are rounded down, so the exact concentration that can be calculated is not exactly 2 in neither (l/r) case, just close to 2.
     
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