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Chemistry question

  1. Aug 23, 2013 #1
    I will ask the teacher personally for further explanation (he was absent), but I would like to complete the homework prior to doing so, so I can get an idea of how to resolve this problem.

    3 sample of NaOH are found at the scene of a crime. The sample used to kill the canadian goose had a 2.0M (I assume its moles) concentration

    suspect had 1 with her 160g of a 200 ml solution of soln of NaOH
    suspect 2 500g of a 500 ml soln of NaOH
    suspect 3 40 g of a 1L soln of NaOH

    purpose? (not to find the killer)
    material
    procedure
    data
    conclusion

    my hypothesis is that it is suspect 3

    23+16+1= 40 g/mol (molarmass) * 2 mol of the sample = 80 g

    first is only lacking 40g, not enough to kill,
    second has all the grams
    3rd lacks a lot of grams, maybe he didnt have the bottle full of the NaOH
     
  2. jcsd
  3. Aug 23, 2013 #2

    NascentOxygen

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    Staff: Mentor

    Hi chemistryhelps, [Broken]

    Don't you mean that when 200 ml of the solution was analyzed it was found to contain 160g of NaOH solid? If so, what molarity would that make it?

    Was there a fourth sample at the scene, one of 2M and which allowed investigators to decide that was the concentration used to kill the bird?
     
    Last edited by a moderator: May 6, 2017
  4. Aug 24, 2013 #3

    chemisttree

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    Science Advisor
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    Gold Member

    What an unusual question! I would start by calculating the concentrations of all three suspects. BTW, 2.0 M is two molar, not 2 moles.
     
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