# Chemistry question

1. Aug 23, 2013

### chemistryhelps

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. Aug 23, 2013

### 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
3. Aug 24, 2013

### chemisttree

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.