Question

A chemist prepares a solution of iron chloride by measuring out 0.10 g of FeCl2 into a 50. mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of anions in the chemist's solution.

Answer 1

Answer: The molarity of anions $(Cl^-)$ in the solution is 0.0316 M

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of iron (II) chloride = 0.10 g

Molar mass of iron (II) chloride = 126.75 g/mol

Volume of solution = 50 mL

Putting values in above equation, we get:

$\text{Molarity of iron (II) chloride}=\frac{0.10\times 1000}{126.75\times 50}\\\\\text{Molarity of iron (II) chloride}=0.0158M$

1 mole of iron (II) chloride produces 1 mole of $Fe^{2+}$ ions and 2 moles of $Cl^-$ ions

So, concentration of chloride ions (anions) in the solution = $(2\times 0.0158)=0.0316M$

Hence, the molarity of anions $(Cl^-)$ in the solution is 0.0316 M