Calculating the chloride concentration
The first thing we need to do is address your chemical equations.  $\ce{MgCl2}$ dissociates in water as follows:  

$$\ce{MgCl2 -&gt; Mg^{2+} + 2Cl-}$$  

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