Question

a chemist adds 330.0 ml of a 2.84×10^{-5} mm mercury(ii) iodide (hgi_{2}) solution to a reaction flask. calculate the micromoles of mercury(ii) iodide the chemist has added to the flask. round your answer to 3 significant digits

Explanation:

Step1: Convert volume to liters

Volume \( V = 330.0\space mL=330.0\times10^{- 3}\space L = 0.3300\space L\)

Step2: Convert concentration to mol/L

Concentration \( C = 2.84\times10^{-5}\space mM = 2.84\times10^{-5}\times10^{-3}\space mol/L=2.84\times10^{-8}\space mol/L\)

Step3: Calculate moles using \(n = C\times V\)

\(n=(2.84\times 10^{-8}\space mol/L)\times(0.3300\space L)=9.372\times10^{-9}\space mol\)

Step4: Convert moles to micromoles

Since \(1\space mol = 10^{6}\space\mu mol\), then \(n = 9.372\times10^{-9}\space mol\times10^{6}\space\mu mol/mol = 9.372\times10^{-3}\space\mu mol\approx9.37\times10^{-3}\space\mu mol\)

Answer:

\(9.37\times 10^{-3}\)