Question #ba500

1 Answer
Jul 26, 2016

#"% v/v" = 20%#

Explanation:

The problem provides you with the volume of solute, which in your case is ethanol, #"CH"_3"CH"_2"OH"#, and the volume of solvent, which in your case is water, so right from the start you know that you're looking for the solution's volume by volume percent concentration, #"% v/v"#.

The solution's volume by volume percent concentration tells you how many cubic centimeters of solute you get for every #"100 cm"^3# of solution. The first thing to do here will be to figure out the total volume of the solution.

#V_"total" = V_"solute" + V_"solvent"#

In your case, you have

#V_"total" = "40 cm"^3 + "160 cm"^3 = "200 cm"^3#

Now, you know that you have #"40 cm"^3# of ethanol in #"200 cm"^3# of solution, which means that #"100 cm"^3# of solution will contain

#100 color(red)(cancel(color(black)("cm"^3color(white)(a)"solution"))) * ("40 cm"^3color(white)(a)"CH"_3"CH"_2"OH")/(200color(red)(cancel(color(black)("cm"^3color(white)(a)"solution")))) = "20 cm"^3color(white)(a)"CH"_3"CH"_2"OH"#

So, if #"100 cm"^3# of solution contain #"20 cm"^3# of ethanol, it follows that the solution's volume by volume percent concentration is

#"% v/v" = color(green)(|bar(ul(color(white)(a/a)color(black)("20% CH"_3"CH"_2"OH")color(white)(a/a)|)))#