Question

A chemist prepares a solution of mercury(II) iodide HgI2 by measuring out 0.0067μmol of mercury(II) iodide into a 350.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /μmolL of the chemist's mercury(II) iodide solution. Round your answer to 2 significant digits.

Answer 1

The concentration of the chemist's mercury(II) iodide solution is 0.019 μmol/L after rounding to two significant digits.

Step-by-step explanation:

The concentration of the chemist's mercury(II) iodide solution can be calculated using the formula for molarity, which is the number of moles of solute divided by the volume of the solution in liters.

In this case, we are given 0.0067μmol of HgI2 and a volume of 0.350 L:

Molarity (μM) = amount of solute (in micromoles) / volume of solution (in liters)

Molarity (μM) = 0.0067μmol / 0.350 L = 0.01914μmol/L

After rounding to two significant digits, the concentration of the solution is 0.019 μM.

Answer 2

`\large\boxed{0.019\, \mu \text{mol/L}}`

Step-by-step explanation:

`\text{Concentration } = \frac{\text{moles}}{\text{litres}}\\\\\text{c} = (n)/(V)`

1. Convert millilitres to litres

`\text{V = 350. mL} * \frac{\text{1 L}}{\text{1000 mL}} = \text{0.3500 L}`

2. Calculate the concentration

`c = \frac{0.0067\, \mu\text{mol}}{\text{0.3500 L}} = \mathbf{0.019 \,  \mu}\textbf{mol/L}\\\\\text{The concentration of mercury(II) iodide is $\large\boxed{\mathbf{0.019\, \mu} \textbf{mol/L}}$}.`