In this case, a
Now, solutions are homogeneous mixtures, which implies that they have the same composition throughout. Consequently, you can say that
In other words, you can use the molarity of the solution as a conversion factor to determine the volume of the solution that must contain
#4.00 color(red)(cancel(color(black)("moles solute"))) * "1 L solution"/(2.8color(red)(cancel(color(black)("moles solute")))) = color(darkgreen)(ul(color(black)("1.4 L solution")))#
The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the solution.