Question

a chemist prepares a solution of sodium chloride (nacl) by measuring out 65.6 mg of nacl into a 250. ml volumetric flask and filling to the mark with distilled water.
calculate the molarity of cl^- anions in the chemists solution.
be sure your answer has the correct number of significant digits.

Explanation:

Step1: Convert the mass of NaCl to moles

The molar mass of \(NaCl\) is \(M = 58.44\space g/mol\). The mass of \(NaCl\) is \(m=65.6\space mg = 65.6\times10^{- 3}\space g\).
Using the formula \(n=\frac{m}{M}\), we have \(n=\frac{65.6\times 10^{-3}\space g}{58.44\space g/mol}\approx1.12\times10^{-3}\space mol\).

Step2: Determine the volume of the solution

The volume of the solution \(V = 250\space mL=250\times10^{-3}\space L = 0.250\space L\).

Step3: Calculate the molarity of \(NaCl\)

Using the formula \(C=\frac{n}{V}\), the molarity of \(NaCl\) is \(C_{NaCl}=\frac{1.12\times 10^{-3}\space mol}{0.250\space L}=4.48\times10^{-3}\space mol/L\).

Step4: Relate the molarity of \(NaCl\) to the molarity of \(Cl^{-}\)

Since \(NaCl = Na^{+}+Cl^{-}\), the molar ratio of \(NaCl\) to \(Cl^{-}\) is \(1:1\). So \(C_{Cl^{-}}=C_{NaCl}\).

Answer:

\(4.48\times 10^{-3}\space mol/L\)