# Question 3d61b

Apr 3, 2015

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 $0.21 M \times 0.0655 L = 0.01376 m o l$.

Ionization of each AlF unit produces one fluoride ion (${F}^{-}$), so the number of moles of fluoride ions is the same, $0.01376 m o l$. As the other contributors have pointed out, the correct formula for aluminum fluoride is $A l {F}_{3}$, and in this case, each formula unit would produce three ${F}^{-}$ ions.

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

$0.01376 m o l \times 6.022 \times {10}^{23} m o {l}^{- 1} = 8.28 \times {10}^{21}$

or, in the case of $A l {F}_{3}$,

$3 \cdot 0.01376 m o l \times 6.022 \times {10}^{23} m o {l}^{- 1} = 2.48 \times {10}^{22}$

Apr 3, 2015

I assume you are referring to an aluminium fluoride, $A l {F}_{3}$, solution.

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 $A l {F}_{3}$ solution, and its volume, you can determine the number of moles you have by

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{A l {F}_{3}} = 0.21 \text{mol"/cancel("L") * 65.5 * 10^(-3)cancel("L") = "0.013755 moles }$ $A l {F}_{3}$

Now, because aluminium fluoride is an ionic compound that's soluble in water, your solution will contain $A {l}^{3 +}$ cations and ${F}^{-}$ anions.

$A l {F}_{3 \left(a q\right)} \to A {l}_{\left(a q\right)}^{3 +} + \textcolor{red}{3} {F}_{\left(a q\right)}^{-}$

Notice that you have a $1 : \textcolor{red}{3}$ mole ratio between aluminium fluoride and the fluoride anions; this means that, for every mole of the former, you'll have 3 times more moles of the latter.

0.013755cancel("moles AlF"_3) * ("3 moles F"^(-))/(1cancel("mole AlF"_3)) = "0.041265 moles F"^(-)

To get the exact number of fluoride ions present in your solution, use the fact that 1 mole of a substance contains exactly $6.022 \cdot {10}^{23}$ molecules of that substance - this is known as Avogadro's number.

In your case, 1 mole of fluoride ions will contain $6.022 \cdot {10}^{23}$ fluoride ions, which means that you have

0.041265cancel("moles F"^(-)) * (6.022 * 10^(23)"F"^(-)"ions")/(1 cancel("mole F"^(-))) = 2.485 * 10^(22)"F"^(-)"ions"

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

$\text{number of F"^(-)"ions} = \textcolor{g r e e n}{2.5 \cdot {10}^{22}}$

Apr 3, 2015

There are $2.5 \times {10}^{22}$ fluoride ions in the $0.21 \text{M AlF"_3}$ solution.

First of all, the formula $\text{AlF}$ is incorrect, so that could be your problem. It should be $\text{AlF"_3}$.

Molarity means moles/L. So $\text{0.21 M}$ is $\text{0.21 mol/L}$.

Convert the volume of the solution from mL to L.

65.5 cancel ("mL)" x $\left(1 \text{L")/(1000 cancel ("mL}\right)$ = $\text{0.0655 L}$

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

0.0655 cancel (L)" x ${\text{0.21 mol AlF}}_{3} / \cancel{L}$ = $\text{0.0137 mol AlF"_3}$

$\text{1 mol AlF"_3}$ contains 3 moles of fluoride ions ($\text{F"^-}$).

$0.0137 \cancel{{\text{mol AlF}}_{3}}$ x (3 "mol F"^-)/(1 cancel ("mol AlF"_3)"# = ${\text{0.0411 mol F}}^{-}$

There are $\text{6.022 x 10"^23}$ ions in one mole of ions.
Multiply mol ${\text{F}}^{-}$ times $6.022 \times {10}^{23} \text{ions/mole}$.

$0.0411 \cancel{\text{mol F"^(-)}}$ x $\left(6.022 \times {10}^{23} {\text{F"^(-))/(1 cancel ("mol F}}^{-}\right)$= $2.48 \times {10}^{22} {\text{F}}^{-}$ = $2.5 \times {10}^{22} {\text{F}}^{-}$ (rounded to two significant figures due to 0.21 M)