Question

You need to make an aqueous solution of 0.173 M zinc fluoride for an experiment in lab, using a 500 mL volumetric flask. How much solid zinc fluoride should you add

Answer 1

About 8.94.

Step-by-step explanation:

Because we are given a 500. mL volumetric flask, the solution will have a volume of 500. mL.

Find the number of moles of zinc fluoride needed. Recall that molarity is simply moles per liter of solution:

`\displaystyle 500.\text{ mL} \cdot \frac{0.173\text{ mol ZnF$_2$}}{1\text{ L}} \cdot \frac{1\text{ L}}{1000\text{ mL}} = 0.0865\text{ mol ZnF$_2$}`

Convert this to grams. The molecular weight of zinc fluoride is 103.38 g/mol:

`\displaystyle 0.0865\text{ mol ZnF$_2$} \cdot \frac{103.38\text{ g ZnF$_2$}}{1\text{ mol ZnF$_2$}} = 8.94\text{ g ZnF$_2$}`

In conclusion, about 8.94 grams of solid zinc fluoride should be added.