Question

Excess of $NaOH_{(aq)}$ was added to $100\, mL$ of ${ FeCl_3}$ (aq) resulting into $2.14\, g$ of ${Fe(OH)_3}$. The molarity of ${FeCl_3}$ (aq) is : (Given molar mass of ${Fe = 56 \, g \, mol^{-1}}$ and molar mass of ${Cl = 35.5 \, g \, mol^{-1}}$)

• A. $0.2\, M$
• B. $0.3\, M$
• C. $0.6\, M$
• D. $1.8\, M$

Answer 1

The Correct Option is A

 To find the molarity of \({FeCl_3}\) (aq), we first need to understand the reaction that takes place when excess \({NaOH}\) is added to \({FeCl_3}\). The reaction can be represented as:

\({FeCl_3}_{(aq)} + 3NaOH_{(aq)} \rightarrow {Fe(OH)_3} \downarrow + 3NaCl_{(aq)}\)

Given that \({2.14 \, g}\) of \({Fe(OH)_3}\) is produced, we first find the moles of \({Fe(OH)_3}\) using its molar mass.

1. Calculate the molar mass of \({Fe(OH)_3}\):
   • Molar mass of \({Fe}\) = \({56 \, g \, mol^{-1}}\)
   • Molar mass of \({O}\) = \({16 \, g \, mol^{-1}}\)
   • Molar mass of \({H}\) = \({1 \, g \, mol^{-1}}\)
   • Total molar mass: \({56 + 3 \times (16 + 1) = 107 \, g \, mol^{-1}}\)
2. Determine the moles of \({Fe(OH)_3}\):
3. Relate the moles of \({FeCl_3}\) to \({Fe(OH)_3}\) using the balanced equation:
4. If \({0.02 \, mol}\) of \({Fe(OH)_3}\) is produced, then \({0.02 \, mol}\) of \({FeCl_3}\) must have been present in the solution.

Calculate the molarity of \({FeCl_3}\):

• Volume of the \({FeCl_3}\) solution = \({100 \, mL = 0.1 \, L}\)
• Molarity, \(M = \frac{\text{moles of } {FeCl_3}}{\text{volume of solution in L}} = \frac{0.02 \, mol}{0.1 \, L} = 0.2 \, M\)

Therefore, the molarity of the \({FeCl_3}\) solution is \({0.2 \, M}\). Hence, the correct answer is \({0.2 \, M}\).