Answer: The molarity of anions [tex](Cl^-)[/tex] in the solution is 0.0316 M
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
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:
[tex]\text{Molarity of iron (II) chloride}=\frac{0.10\times 1000}{126.75\times 50}\\\\\text{Molarity of iron (II) chloride}=0.0158M[/tex]
1 mole of iron (II) chloride produces 1 mole of [tex]Fe^{2+}[/tex] ions and 2 moles of [tex]Cl^-[/tex] ions
So, concentration of chloride ions (anions) in the solution = [tex](2\times 0.0158)=0.0316M[/tex]
Hence, the molarity of anions [tex](Cl^-)[/tex] in the solution is 0.0316 M
