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

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.