Answer:
The concentration of Cl in the solution is 1 M (option E)
Step-by-step explanation:
MgCl₂ is an ionic compound, where subscript 2 indicates that it contributes 2 anions (negatively charged ions) Cl- to the solution. And you know that 0.20 moles of MgCl₂ are dissolved in water. Then, by multiplying this amount by the subscript, you get the amount of the chlorine ion Cl- that contributes the MgCl₂ compound to the solution. Then it is obtained that said compound contributes 0.40 moles of the Cl- anion.
With a thought similar to the previous case, it can be seen that in this case 0.10 mol of the ionic compound KCl is dissolved in water, and that this compound in turn contributes only a Cl- anion to the solution. Then the KCl compound provides 0.10 moles of the Cl- anion.
On the other hand, molarity is the number of moles of solute that are dissolved in a given volume. It is expressed in the units (moles/liter).
Then, the volume in which the compounds are dissolved in the corresponding unit of measure must be expressed, that is to say in L. Knowing that 1 L is 1000 mL, then 500 mL represents 0.50 L.
And knowing that the MgCl₂ compound contributes 0.40 moles of the Cl- anion while the KCl compound contributes 0.10 moles of the anion, it is possible to say that in total there are 0.50 moles of the Clons present anion in the solution.
So:
concentration of Cl= 1 mol/L= 1 M
The concentration of Cl in the solution is 1 M (option E)