Question

A chemist prepares a solution of mercury(II) iodide by weighing out of mercury(II) iodide into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's mercury(II) iodide solution. Be sure your answer has the correct number of significant digits.

Answer 1

The question is incomplete, here is the complete question:

A chemist prepares a solution of mercury(II) iodide
`(HgI_2)` 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
`3.05* 10^(-8)mol/L`

Step-by-step explanation:

To calculate the molarity of solution, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in mL)}}`

We are given:

Moles of mercury (II) iodide =
`0.0122\mu mol=0.0122* 10^(-6)mol` (Conversion factor:
`1mol=10^6\mu mol` )

Volume of solution = 400. mL

Putting values in above equation, we get:

`\text{Molarity of }HgI_2=(0.0122* 10^(-6)* 1000)/(400.)\\\\\text{Molarity of }HgI_2=3.05* 10^(-8)mol/L`

Hence, the molarity of chemist's mercury (II) iodide solution is
`3.05* 10^(-8)mol/L`