Solve Chemistry Problem: 3 NaOH Samples Found at Crime Scene

[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

 

[NascentOxygen]

suspect had 1 with her 160g of a 200 ml solution of soln of NaOH

Hi chemistryhelps,

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?

 

[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.