# How many grams of MgCl_2 would be required to produce a 4.5 M solution with a volume of 2.0 L?

Jun 7, 2017

${\text{860 g MgCl}}_{2}$

#### Explanation:

For starters, we know that a $\text{1-M}$ solution contains $1$ mole of solute for every $\text{1.0 L}$ of solution.

This implies that a $\text{4.5-M}$ magnesium chloride solution will contain $4.5$ moles of magnesium chloride, the solute, for every $\text{1.0 L}$ of solution.

So all you have to do now is figure out how many moles of magnesium chloride would be equivalent in $\text{2.0 L}$ of solution to $4.5$ moles in '1.0 L" of solution.

(color(red)(?)color(white)(.)"moles MgCl"_2)/"2.0 L solution" = overbrace("4.5 moles MgCl"_2/("1.0 L solution"))^(color(blue)("= 4.5 M solution"))

Use cross multiplication to get

color(red)(?) = (2.0 color(red)(cancel(color(black)("L solution"))))/(1.0color(red)(cancel(color(black)("L solution")))) * "4.5 moles MgCl"_2

This will be equal to

color(red)(?) = "9.0 moles MgCl"_2

So, you know that you can get a $\text{4.5-M}$ magnesium chloride solution by dissolving $9.0$ moles of magnesium chloride in enough water to have a total volume of $\text{2.0 L}$ of solution.

To convert this to grams, use the compound's molar mass

$9.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles MgCl"_2))) * "95.211 g"/(1color(red)(cancel(color(black)("mole MgCl"_2)))) = color(darkgreen)(ul(color(black)("860 g}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for your values.