The question is incomplete, here is the complete question:
A chemist prepares a solution of mercury(II) iodide [tex](HgI_2)[/tex] by measuring out 0.0122 µmol of mercury(II) iodide into a 400 mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's mercury(II) iodide solution. Be sure your answer has the correct number of significant digits.
Answer: The molarity of chemist's mercury (II) iodide solution is [tex]3.05\times 10^{-8}mol/L[/tex]
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
We are given:
Moles of mercury (II) iodide = [tex]0.0122\mu mol=0.0122\times 10^{-6}mol[/tex] (Conversion factor: [tex]1mol=10^6\mu mol[/tex] )
Volume of solution = 400. mL
Putting values in above equation, we get:
[tex]\text{Molarity of }HgI_2=\frac{0.0122\times 10^{-6}\times 1000}{400.}\\\\\text{Molarity of }HgI_2=3.05\times 10^{-8}mol/L[/tex]
Hence, the molarity of chemist's mercury (II) iodide solution is [tex]3.05\times 10^{-8}mol/L[/tex]
