# Calculating the chloride concentration

What are the concentrations of $\ce{Cl-}$ in the following solution? $1.9\ \mathrm{g}\ \ce{MgCl2}$ is dissolved in water to make $1\ \mathrm{L}$ total solution

$$48.31\ \mathrm{g/mol}\ \ce{MgCl2}$$

$$(1.9~\mathrm{g}\ \ce{MgCl2})(1\ \mathrm{mol}/48.31\ \mathrm{g}) = 3.932\times 10^{-2}\ \mathrm{mol}\ \ce{MgCl2}$$

$$\ce{MgCl2 -> Mg+ + Cl2-}$$

$$(3.932\times 10^{-2}\ \mathrm{mol}\ \ce{MgCl2})((1\ \mathrm{mol}\ \ce{Cl2})/(1\ \mathrm{mol}\ \ce{MgCl2})) = 3.932\times 10^{-2}\ \mathrm{mol}\ \ce{Cl2-}$$

then

$$(3.932\times 10^{-2}\ \mathrm{mol}\ \ce{Cl2-})/(1~\mathrm{L}) = 3.932\times 10^{-2}\ \mathrm{M}\ \ce{Cl2-}$$

Is that correct? Or does the $2$ go infront of the $\ce{Cl}$, so the mole ratio is $((2\ \mathrm{mol}\ \ce{Cl-})/(1\ \mathrm{mol}\ \ce{MgCl2}))$?

The first thing we need to do is address your chemical equations. $\ce{MgCl2}$ dissociates in water as follows:
$$\ce{MgCl2 -> Mg^{2+} + 2Cl-}$$
Now you can see that one mole of $\ce{MgCl2}$ dissociates to give two moles of $\ce{Cl-}$. So all we have to do then is calculate the number of moles of $\ce{MgCl2}$ from the given mass and its molecular weight, then multiply that by 2 to get the number of moles of $\ce{Cl-}$:
$$\mathrm{\frac{1.9\ g\ \ce{MgCl2}}{ 95.2\ \frac{g\ \ce{MgCl2}}{mol\ \ce{MgCl2}} } = 0.0200\ mol\ \ce{MgCl2}}$$
$$\mathrm{0.0200\ mol\ \ce{MgCl2}\ *\frac{2\ mol\ \ce{Cl-}}{mol\ \ce{MgCl2}} =0.0400\ mol\ \ce{Cl-}}$$
Then since the volume is 1L, the $\ce{Cl-}$ concentration is just $\mathrm{0.040 \frac{mol}{L}}$.