[MustangFTW]

Im stuck on this question, i keep getting the wrong answer.



Determine the concentration of chloride atoms in a solution that is prepared by diluting 3.75 mL of .228 M MgCl2 to a final volume of 50 mL

no matter how i try to solve the problem i end up with .0171 M, which isnt the right answer, answer key says it should be .0342 M.

The way i did it was:

.228 = mols/.00375L
mols = 8.55e-4
M=8.55e-4/.05L
M=.0171

[MustangFTW]

Nevermind i got it:


the equation:

Mass of element = mass of sample x (mass of element in 1 mol of compound/mass of 1 mol of compound)

element is the keyword i guess, using the answer from this equation, which is in grams, divide by the mass of one mole of the element to get moles of the element, then divide by .05L to get .0342.


Hope this helps someone haha.

[ramboacid]

Probably an easier way to do that is to notice that for every formula unit of MgCl₂, one Mg²⁺ ion and two Cl^- ions would be formed, according to the equation
MgCl₂ Mg²⁺ + 2Cl^-

The mole ratio of Cl^- to MgCl₂ is 2:1. Therefore, the .228 M MgCl₂ solution is actually .456 M in Cl^- (.228 times 2). Then you can proceed to calculate the molarity of the diluted solution from there.

Also, just for future reference, chloride ions are formed, not chlorine atoms, because the chlorine atoms carry an extra electron and hence have a -1 charge.