Answer: The molarity of chloride anions in the solution is 0.01344 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]
Given mass of ferric chloride = 72.7 mg = 0.0727 g (Conversion factor: 1 g = 1000 mg)
Molar mass of ferric chloride = 162.2 g/mol
Volume of solution = 100 mL
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{0.0727\times 1000}{162.2\times 100}\\\\\text{Molarity of solution}=0.00448M[/tex]
1 mole of ferric chloride contains 1 mole of [tex]Fe^{3+}[/tex] ions and 3 moles of [tex]Cl^-[/tex] ions
Moles of [tex]Cl^-[/tex] = (3 × 0.00448) = 0.01344 M
Hence, the molarity of chloride anions in the solution is 0.01344 M
