Question

A chemist prepares a solution of ironIII chloride FeCl3 by measuring out 72.7mg of FeCl3 into a 100.mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of Cl− anions in the chemist's solution.

Answer 1

Answer: The molarity of chloride anions in the solution is 0.01344 M

Explanation:

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

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

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:

$\text{Molarity of solution}=\frac{0.0727\times 1000}{162.2\times 100}\\\\\text{Molarity of solution}=0.00448M$

1 mole of ferric chloride contains 1 mole of $Fe^{3+}$ ions and 3 moles of $Cl^-$ ions

Moles of $Cl^-$ = (3 × 0.00448) = 0.01344 M

Hence, the molarity of chloride anions in the solution is 0.01344 M