How many liters of 4.00 M solution can be made using exactly 100.0 grams of lithum bromide?

The first thing that you need to do here is to convert the mass of lithium bromide to moles by using the compound's molar mass.

#100.0 color(red)(cancel(color(black)("g"))) * "1 mole LiBr"/(86.845color(red)(cancel(color(black)("g")))) = "1.1515 moles LiBr"#

Now, the molarity of the solution is simply a measure of the number of moles of solute, which in your case is lithium bromide, present for every #"1.00 L"# of the solution.

In order to have a #"4.00-M"# solution of lithium bromide, you need to have #4.00# moles of lithium bromide for every #"1.00 L"# of this solution.

You know that your sample contains #1.1515# moles of lithium bromide, so you can use the molarity of the solution as a conversion factor to determine how many liters of this solution can be made.

#1.1515 color(red)(cancel(color(black)("moles LiBr"))) * "1.00 L solution"/(4.00color(red)(cancel(color(black)("moles LiBr")))) = color(darkgreen)(ul(color(black)("0.288 L solution")))#

The answer is rounded to three sig figs, the number of sig figs you have for the molarity of the solution.