[sauganash]

Here's the homework question I'm having difficulties with:

What is the concentration of AlCl3 solution if 150 ml of the solution contains 650 mg of Cl ion?

Here's my work:
If the [Cl-]= 650 mg, the AlCl3 is 1/3 of that, since 1 mol of AlCl3 produces 3 moles of Cl- ions

Concentration (Molarity)  of Cl-ion= moles of solute / Liters of solution
Moles of Cl ion:

650 mg Cl * 1 g       *   1 mole Cl
                   1000 mg        35.45 g

= 0.1833 moles of Cl ion

Liters of solution:
150 ml = 0.15 L

C = 0.1833 moles Cl / 0.15 L
=0.122 M

However, when the solution disassociates in the solution:
AlCl3  Al + 3Cl
If Al/Cl = x, then
AlCl3  x + 3x

Concentration of solution:
AlCl3 = (x)(3x) = 3x2
=3(.122)2 = .0446 M


However, my answer does not match any of teh answer choices. Please *delete me*

[enahs]

You are good until your last little bit, then it just wacky.

You are right, for every 3 mols of Cl- ion in solution you had 1 mol AlCl₃.

So what is the concentration of the Cl- ion? The AlCl₃ concentration is just 1/3rd that...no?


# Cl- Ion    *    1 AlCl₃  =  ??
                      3 Cl- Ion

[Borek]

650 mg Cl * 1 g       *   1 mole Cl
                   1000 mg        35.45 g

= 0.1833 moles of Cl ion

Somehow you managed to start OK, but to end wrong.