How many liters of a 30% alcohol solution must be mixed with 20 liters of a 80% solution to get a 70% solution?

1 Answer
Aug 2, 2017

Answer:

#5# #"L"#

Explanation:

I'll assume the given percentages are those by volume (i.e. #"v/v"#)..

We're asked to find the volume, in #"L"#, of a #30%# #"v/v"# alcohol solution must be added to #20# #"L"# of a #80%# #"v/v"# to obtain a solution that is #70%# #"v/v"#.

To do this, we can set up sort of an algebraic equation that represents this situation.

Here, we have

  • #x# liters of #0.3# fractional volume solution

  • #20# liters of #0.8# fractional volume solution

These two added together must be a solution with fractional volume #0.7# (the volume will be #20# #"L"# #+ x#), so we can write an equation

#ul(overbrace(x)^"volume of 30%"(0.3) + overbrace((20))^"volume of 80%"(0.8) = overbrace((x+20))^"total volume"(0.7)#

Now, we simply solve for #x#:

#0.3x + 16 = 0.7x + 14#

#0.4x = 2#

#color(red)(ul(x = 5#

So the required volume of the #30%# by volume solution is

#color(red)(ulbar(|stackrel(" ")(" "5color(white)(l)"L"" ")|)#