Questions: A chemist prepares a solution of iron(II) chloride (FeCl2) by measuring out 1.50 g of FeCl2 into a 300 mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of Cl- anions in the chemist's solution. Be sure your answer is rounded to 3 significant digits. mol/L × 10^5

A chemist prepares a solution of iron(II) chloride (FeCl2) by measuring out 1.50 g of FeCl2 into a 300 mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of Cl- anions in the chemist's solution. Be sure your answer is rounded to 3 significant digits. mol/L × 10^5
Transcript text: A chemist prepares a solution of iron(II) chloride $\left(\mathrm{FeCl}_{2}\right)$ by measuring out 1.50 g of $\mathrm{FeCl}_{2}$ into a 300 . mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of $\mathrm{Cl}^{-}$anions in the chemist's solution. Be sure your answer is rounded to 3 significant digits. $\frac{\mathrm{mol}}{\mathrm{L}}$ $\square \times 10^{5}$
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Solution

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Solution Steps

Step 1: Determine the Molar Mass of FeCl\(_2\)

First, calculate the molar mass of iron(II) chloride, \(\mathrm{FeCl}_2\). The atomic masses are approximately:

  • Fe: 55.845 g/mol
  • Cl: 35.453 g/mol

The molar mass of \(\mathrm{FeCl}_2\) is: \[ 55.845 + 2 \times 35.453 = 126.751 \, \text{g/mol} \]

Step 2: Calculate Moles of FeCl\(_2\)

Next, calculate the number of moles of \(\mathrm{FeCl}_2\) in 1.50 g: \[ \text{moles of } \mathrm{FeCl}_2 = \frac{1.50 \, \text{g}}{126.751 \, \text{g/mol}} = 0.01183 \, \text{mol} \]

Step 3: Determine Moles of Cl\(^-\) Ions

Each \(\mathrm{FeCl}_2\) molecule contains 2 chloride ions. Therefore, the moles of \(\mathrm{Cl}^-\) ions are: \[ 0.01183 \, \text{mol} \times 2 = 0.02366 \, \text{mol} \]

Step 4: Calculate Molarity of Cl\(^-\) Ions

The volume of the solution is 300 mL, which is 0.300 L. The molarity of \(\mathrm{Cl}^-\) ions is: \[ \text{Molarity of } \mathrm{Cl}^- = \frac{0.02366 \, \text{mol}}{0.300 \, \text{L}} = 0.07887 \, \text{mol/L} \]

Final Answer

The molarity of \(\mathrm{Cl}^-\) ions in the solution, rounded to three significant digits, is: \[ \boxed{7.89 \times 10^{-2} \, \text{mol/L}} \]

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