Question #6cd19

As you know, one formula unit of sodium chloride, `"NaCl"`, contains

• one sodium cation, `1 xx "Na"^(+)`
• one chloride anion, `1 xx "Cl"^(-)`

On the other hand, one formula unit of magnesium chloride, `"MgCl"_2`, contains

• one magnesium cation, `1 xx "Mg"^(2+)`
• two chloride anions, `2 xx "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 `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 `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.