Question #5f823

1 Answer
Jun 17, 2016

Answer:

#"0.180 M"#

Explanation:

The idea here is that you need to use the underlying principle of a dilution, which tells you that the concentration of a solution can be decreased by

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

http://acidsandbasesfordummieschem.weebly.com/molarity.html

Mathematically, this can be expressed by using the molarities and volumes of the stock solution, which is the solution you're starting with, and the of the target solution

#color(blue)(overbrace(c_1 xx V_1)^(color(darkgreen)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(darkgreen)("moles of solute in diluted solution"))#

Here

#c_1#, #V_1# - the molarity and volume of the concentrated solution
#c_2#, #V_2# - the molarity and volume of the diluted solution

So, your starting solution has a volume of #"100.0 mL"# and a molarity of #"1.35 M"#

#{(V_1 = "100.0 mL"), (c_1 = "1.35 M"):}#

The volume of the target solution will include the added water

#V_2 = "100.0 mL" + "650.0 mL" = "750.0 mL"#

Rearrange the above equation to solve for #c_2#, the concentration of the target solution, i.e. of the diluted solution.

#c_1V_1 = c_2V_2 implies c_2 = V_1/V_2 * c_1#

Plug in your values to find

#c_2 = (100.0 color(red)(cancel(color(black)("mL"))))/(750.0color(red)(cancel(color(black)("mL")))) * "1.35 M" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.180 M")color(white)(a/a)|)))#

The answer is rounded to three sig figs.