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

##### 1 Answer

#### Explanation:

For starters, we know that a **mole** of solute **for every** **of solution**.

This implies that a **moles** of magnesium chloride, the solute, for every

So all you have to do now is figure out how many moles of magnesium chloride would be **equivalent** in **moles** in

#(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 **moles** of magnesium chloride in enough water to have a total volume of

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

#9.0 color(red)(cancel(color(black)("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.