Question #12ca1

1 Answer
Mar 30, 2016

Answer:

Here's what I got.

Explanation:

The underlying principle of a dilution is that you can decrease the concentration of the solution by

  • keeping the number of moles of solute constant
  • increasing the total volume of the solution

So, in order to dilute a solution you must increase its volume without changing the amount of solute it contains.

Let's assume that solution #"A"# contains an unknown number of moles of solute, #n_A#. The initial concentration of the solution can be written as

#["A"]_0 = n_A/V_(A0)#

#["A"]_0 = n_A/(0.3 * 10^(-3)"L") = (10/3 * 10^3 * n_A)color(white)(a)"mol L"^(-1)#

After you mix this solution with solution #"B"#, the total volume of the resulting solution will be

#V_"total" = "0.3 mL" + "6.7 mL" = "7.0 mL"#

Keeping in mind the fact tha the resulting solution must contain #n_A# moles of solute #"A"#, you an say that its new concentration is equal to

#["A"]_"dil" = n_A/(7.0 * 10^(-3)"L") = (1/7 * 10^3 * n_A)color(white)(a)"mol L"^(-1)#

So, by what factor, #"D.F"#, was solution #"A"# diluted? Divide the initial concentration by the final concentration to get

#"D.F." = ( 10/3 * color(blue)(cancel(color(black)(10^3 * n_a))) color(red)(cancel(color(black)("mol L"^(-1)))))/(1/7 * color(blue)(cancel(color(black)(10^3 * n_a))) color(red)(cancel(color(black)("mol L"^(-1)))))#

#"D.F." = 10/3 * 7 = color(green)(bar(ul(|color(white)(a/a)23.3color(white)(a/a)|)))#

This tells you that the resulting solution is #23.3# times less concentrated than solution #"A"# compared with the initial concentration of #"A"#.

An interesting thing to notice here is that you can get the dilution factor by dividing the final volume by the initial volume of the solution

#"D.F." = (7.0 color(red)(cancel(color(black)("mL"))))/(0.3color(red)(cancel(color(black)("mL")))) = color(green)(bar(ul(|color(white)(a/a)23.3color(white)(a/a)|))) -># for solution #"A"#

The exact same approach can be used to find the dilution of #"B"#. This time, the initial volume will be equal to #"6.7 mL"# and the final volume to #"7.0 mL"#

#"D.F." = (7.0color(red)(cancel(color(black)("mL"))))/(6.7color(red)(cancel(color(black)("mL")))) = color(green)(bar(ul(|color(white)(a/a)1.04color(white)(a/a)|))) -># for solution #"B"#

So, to sum this, the dilution factor for a given dilution is equal to

#color(blue)(|bar(ul(color(white)(a/a)"D.F." = "initial concentration"/"final concentration" = "final volume"/"initial volume"color(white)(a/a)|)))#