In your case, the solution is
So regardless of the volume of the solution, you can say for a fact that you will have a ratio of
Now, you can use the solution's volume by volume percent concentration as a conversion factor to calculate the number of grams of isopropyl alcohol present in
#473 color(red)(cancel(color(black)("mL solution"))) * overbrace("70. mL isopropyl alcohol"/(100color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 70.% v/v")) = color(darkgreen)(ul(color(black)("330 mL isopropyl alcohol")))#
The answer is rounded to two sig figs, the number of sig figs you have for the percent concentration of the solution.