# Question 72c75

Sep 30, 2016

Here's what I got.

#### Explanation:

The idea here is that you're dealing with three soluble ionic compounds, which implies that they will dissociate completely in aqueous solution. You will have

${\text{AlCl"_ (3(aq)) -> "Al"_ ((aq))^(3+) + color(blue)(3)"Cl}}_{\left(a q\right)}^{-}$

${\text{MgCl"_ (2(aq)) -> "Mg"_ ((aq))^(2+) + color(purple)(2)"Cl}}_{\left(a q\right)}^{-}$

${\text{NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

Notice that every mole of aluminium chloride produces $\textcolor{b l u e}{3}$ moles of chloride anions in aqueous solution.

Similarly, every mole of magnesium chloride produces $\textcolor{p u r p \le}{2}$ moles of chloride anions and eery mole of sodium chloride produces $1$ mole of chloride anions when dissolved in water.

All you have to do now is use the molarities and volumes of the three solutions to find which one will contain the largest number of moles of chloride anions

• aluminium chloride, ${\text{AlCl}}_{3}$

100.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * (0.30color(red)(cancel(color(black)("moles AlCl"_3))))/(1color(red)(cancel(color(black)("L")))) * (color(blue)(3)color(white)(a)"moles Cl"^(-))/(1color(red)(cancel(color(black)("mole AlCl"_3))))

$= {\text{0.090 moles Cl}}^{-}$

• amgnesium chloride, ${\text{MgCl}}_{2}$

50.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * (0.60color(red)(cancel(color(black)("moles MgCl"_2))))/(1color(red)(cancel(color(black)("L")))) * (color(purple)(2)color(white)(a)"moles Cl"^(-))/(1color(red)(cancel(color(black)("mole MgCl"_2))))

$= {\text{0.060 moles Cl}}^{-}$

• sodium chloride, $\text{NaCl}$

200.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * (0.40color(red)(cancel(color(black)("moles NaCl"))))/(1color(red)(cancel(color(black)("L")))) * "1 mole Cl"^(-)/(1color(red)(cancel(color(black)("mole NaCl"))))#

$= {\text{0.080 moles Cl}}^{-}$

As you can see, the aluminium chloride solution will contain the largest number of moles of chloride anions.

In terms of the number of moles of chloride anions, you can say that you have

${\text{100.0 mL, 0.30 M AlCl"_3 > "200.0 mL, 0.40 M NaCl" > "50.0 mL, 0.60 M MgCl}}_{2}$