# Calculate the number of moles of Cl^(-) ions in 1.75 L of 1.0 xx 10^-3 M AlCl_3. How would you solve this?

May 9, 2017

$5.3 \cdot {10}^{- 3} {\text{moles Cl}}^{-}$

#### Explanation:

For starters, calculate how many moles of aluminium chloride would be needed in order to get a $1.0 \cdot {10}^{- 3}$ $\text{M}$ solution in $\text{1.75 L}$ of solution.

As you know, molarity tells you the number of moles of solute present in $\text{1 L}$ of solution. This means that $\text{1 L}$ of $1.0 \cdot {10}^{- 3}$ $\text{M}$ aluminium chloride solution will contain $1.0 \cdot {10}^{- 3}$ moles of aluminium chloride, the solute.

In your case, the solution will contain

1.75 color(red)(cancel(color(black)("L solution"))) * overbrace((1.0 * 10^(-3)color(white)(.)"moles AlCl"_3)/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)(=1.0 * 10^(-3)"M AlCl"_3)) = 1.75 * 10^(-3) ${\text{moles AlCl}}_{3}$

So, you know that you must dissolve $1.75 \cdot {10}^{- 3}$ moles of aluminium chloride in enough water to make the volume of the solution equal to $\text{1.75 L}$.

Now, aluminium chloride is a soluble ionic compound, which implies that it dissociates completely in aqueous solution to produce aluminium actions and chloride anions

${\text{AlCl"_ (color(red)(3)(aq)) -> "Al"_ ((aq))^(3+) + color(red)(3)"Cl}}^{-}$

Notice that every $1$ mole of aluminium chloride dissolved in water produces $\textcolor{red}{3}$ moles of chloride anions.

This implies that your solution will contain

1.75 * 10^(-3) color(red)(cancel(color(black)("moles AlCl"_3))) * (color(red)(3)color(white)(.)"moles Cl"^(-))/(1color(red)(cancel(color(black)("mole AlCl"_3))))

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{5.3 \cdot {10}^{- 3} \textcolor{w h i t e}{.} {\text{moles Cl}}^{-}}}}$

The answer must be rounded to two sig figs, the number of sig figs you have for the molarity of the solution.