# Question 6cd19

Aug 9, 2016

$\text{263 g}$

#### Explanation:

As you know, one formula unit of sodium chloride, $\text{NaCl}$, contains

• one sodium cation, $1 \times {\text{Na}}^{+}$
• one chloride anion, $1 \times {\text{Cl}}^{-}$

On the other hand, one formula unit of magnesium chloride, ${\text{MgCl}}_{2}$, contains

• one magnesium cation, $1 \times {\text{Mg}}^{2 +}$
• two chloride anions, $2 \times {\text{Cl}}^{-}$

This means that for one mole of sodium chloride you're getting $2$ moles of ions, and for one mole of magnesium chloride you're getting $3$ moles of ions.

In other words, the number of ions present in one mole of magnesium chloride, i.e. $3$ moles of ions, is also present in $\frac{3}{2}$ moles of sodium chloride, since

3/2 color(red)(cancel(color(black)("moles NaCl"))) * "2 moles ions"/(1color(red)(cancel(color(black)("mole NaCl")))) = "3 moles ions"

All you have to do now is calculate how many moles of magnesium chloride you have in your sample

285 color(red)(cancel(color(black)("g NaCl"))) * "1 mole MgCl"_2/((24 + 2 xx 35.5)color(red)(cancel(color(black)("g NaCl")))) = "3 moles MgCl"_2

Since you know that you need $\frac{3}{2}$ moles of sodium chloride to get the same number of ions as $1$ mole of magnesium chloride, you can say that you have

3 color(red)(cancel(color(black)("moles MgCl"_2))) * (3/2color(white)(a) "moles Na")/(1color(red)(cancel(color(black)("mole MgCl"_2)))) = 9/2color(white)(a) "moles NaCl"

Finally, use the molar mass of sodium chloride to convert this to grams

9/2 color(red)(cancel(color(black)("moles NaCl"))) * ((23 + 35.5)color(white)(a)"g")/(1color(red)(cancel(color(black)("mole NaCl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("263 g NaCl")color(white)(a/a)|)))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of magnesium chloride.