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

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

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

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

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


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

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 $\mathrm{0.040 \frac{mol}{L}}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.