Answer:
About 8.94.
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:
[tex]\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$_2$}[/tex]
Convert this to grams. The molecular weight of zinc fluoride is 103.38 g/mol:
[tex]\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$_2$}[/tex]
In conclusion, about 8.94 grams of solid zinc fluoride should be added.
