# Question #3d61b

##### 3 Answers

A 0.21 M solution of AlF contains 0.21 moles of molecules for each liter of solution. In this problem we have only 65.5 mL (0.0655 L) of solution, so the number of moles is

Ionization of each AlF unit produces one fluoride ion (*three*

Finally, to obtain the actual number of ions, we multiply by Avogadro's number:

or, in the case of

I assume you are referring to an aluminium fluoride,

To determine the number of fluoride ions present in that particular solution, you must first determine how many moles of aluminium fluoride you have.

Molarity is defined as the number of mole of solute, in your case aluminium fluoride, divided by liters of solution. Since you know what the molarity of the

Now, because aluminium fluoride is an ionic compound that's soluble in water, your solution will contain

Notice that you have a *every mole of the former*, you'll have **3 times more** moles of the latter.

To get the *exact number* of fluoride ions present in your solution, use the fact that *1 mole* of a substance contains exactly **Avogadro's number**.

In your case, *1 mole* of fluoride ions will contain

Rounded to two sig figs, the number of sig figs given for 0.21 M, the answer will be

There are

First of all, the formula

Molarity means moles/L. So

Convert the volume of the solution from mL to L.

Now multiply the volume of the solution in liters times the molarity.

There are

Multiply mol