Question

a chemist prepares a solution of iron(ii) bromide (febr₂) by measuring out 0.17 g of febr₂ into a 300. ml volumetric flask and filling to the mark with distilled water. calculate the molarity of br⁻ anions in the chemists solution. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Calculate the moles of \(FeBr_2\)

The molar mass of \(FeBr_2\) is \(M = 55.85+(2\times79.90)=215.65\space g/mol\).
Using the formula \(n=\frac{m}{M}\), where \(m = 0.17\space g\) and \(M = 215.65\space g/mol\), we have \(n=\frac{0.17\space g}{215.65\space g/mol}\approx7.88\times10^{-4}\space mol\).

Step2: Determine the moles of \(Br^-\) ions

From the formula \(FeBr_2\), we know that \(1\space mol\space FeBr_2\) dissociates to give \(2\space mol\space Br^-\) ions.
So, \(n(Br^-)=2\times n(FeBr_2)=2\times7.88\times 10^{-4}\space mol = 1.576\times10^{-3}\space mol\).

Step3: Calculate the volume of the solution in liters

The volume \(V = 300\space mL=300\times10^{- 3}\space L = 0.300\space L\).

Step4: Calculate the molarity of \(Br^-\) ions

Using the formula \(M=\frac{n}{V}\), where \(n = 1.576\times10^{-3}\space mol\) and \(V = 0.300\space L\), we have \(M=\frac{1.576\times10^{-3}\space mol}{0.300\space L}\approx5.25\times10^{-3}\space mol/L\).

Answer:

\(5.3\times 10^{-3}\space\frac{mol}{L}\)