Question

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

Answer 1

`"860 g MgCl"_2`

Explanation:

For starters, we know that a `"1-M"` solution contains `1` mole of solute for every `"1.0 L"` of solution.

This implies that a `"4.5-M"` magnesium chloride solution will contain `4.5` moles of magnesium chloride, the solute, for every `"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 `"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 `"4.5-M"` magnesium chloride solution by dissolving `9.0` moles of magnesium chloride in enough water to have a total volume of `"2.0 L"` of solution.

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.