#"750. mL"# of a #"5.0 M"# solution needs to be diluted to #"1.0 M"#. How much water should be added to reach this concentration?

1 Answer
Mar 3, 2017

#3.0 * 10^3"mL"#

Explanation:

The trick here is to realize that because the amount of solute must remain constant in a dilution, the decrease in concentration must be equal to the increase in volume.

In other words, diluting a solution will decrease its concentration by a factor #"DF"# and increase its volume by the same factor #"DF"#.

This factor is called the dilution factor and can be written as

#color(blue)(ul(color(black)("DF" = overbrace(c_"concentrated"/c_"diluted")^(color(red)("decrease in concentration")) = overbrace(V_"diluted"/V_"concentrated")^(color(purple)("increase in volume")) )))#

In your case, the concentration decreases by a factor of

#"DF" = (5.0 color(red)(cancel(color(black)("M"))))/(1.0color(red)(cancel(color(black)("M")))) = color(blue)(5)#

which means that the volume must increase by factor of #color(blue)(5)#

#V_"diluted" = color(blue)(5) * "750. mL" = "3750 mL"#

This means that you must add

#"volume of water" = "3750 mL" - "750. mL" = color(darkgreen)(ul(color(black)(3.0 * 10^(3)color(white)(.)"mL")))#

of water to your concentrated solution in order to dilute it from #"5 M"# to #"1 M"#.

The answer is rounded to two sig figs.