Question

A 0.20 mol sample of MgCl2(s) and a 0.10 mol sample of KCl(s) are dissolved in water and diluted to 500 mL. What is the concentration of Cl- in the solution?

Answer 1

Answer:

1 M

Explanation:

Magnesium chloride will furnish chloride ions as:

$MgCl_2\rightarrow Mg^{2+}+2Cl^-$

Given :

Moles of magnesium chloride = 0.20 mol

Thus, moles of chlorine furnished by magnesium chloride is twice the moles of magnesium chloride as shown below:

$Moles =2\times 0.20\ moles$

Moles of chloride ions by magnesium chloride = 0.40 moles

Potassium chloride will furnish chloride ions as:

$KCl\rightarrow K^{+}+Cl^-$

Given :

Moles of potassium chloride = 0.10 moles

Thus, moles of chlorine furnished by potassium chloride is same as the moles of potassium chloride as shown below:

Moles of chloride ions by potassium chloride = 0.10 moles

Total moles = 0.40 + 0.10 moles = 0.50 moles

Given, Volume = 500 mL = 0.5 L (1 mL = 10⁻³ L)

Concentration of chloride ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{Cl^-}=\frac{0.50}{0.5}$

The final concentration of chloride anion = 1 M