Question

7. Suppose 1.01 g of iron (III) chloride is placed in a 10.00-mL volumetric flask with a bit of water in it. The flask is shaken to dissolve the solid and the flask is then filled to the mark. What is the molarity of the final solution?

Answer 1

Answer: The molarity of Iron (III) chloride is 0.622 M.

Explanation:

Molarity is defined as the number of moles present in one liter of solution.  The equation used to calculate molarity of the solution is:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Or,

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

We are given:

Mass of iron (III) chloride = 1.01 g

Molar mass of iron (III) chloride = 162.2 g/mol

Volume of the solution = 10 mL

Putting values in above equation, we get:

$\text{Molarity of Iron (III) chloride}=\frac{1.01g\times 1000}{162.2g/mol\times 10mL}\\\\\text{Molarity of Iron (III) chloride}=0.622M$

Hence, the molarity of Iron (III) chloride is 0.622 M.