Question #61472

The trick here is to realize that the mole fraction of copper(II) perchlorate is the same regardless of the mass, and consequently, the volume of this solution.

As you know the molarity of the solution tells you the number of moles of solute, which in your case is copper(II) perchlorate, present in #"1 L" = 10^3# #"mL"# of solution.

To make the calculations easier, pick a #10^3"-mL"# sample of this solution, Use its density to find the mass of the solution

#10^3 color(red)(cancel(color(black)("g"))) * "1.15 g"/(1color(red)(cancel(color(black)("g")))) = "1150 g"#

Now, you know that the mole fraction of copper(II) perchlorate, which is defined as the number of moles of copper(II) perchlorate divided by the total number of moles present in the solution, is equal to #0.335#.

If you take #x# to be the number of moles of copper(II) perchlorate and #y# to be the number of moles of water present in this sample, you can say that

#(x color(red)(cancel(color(black)("moles"))))/((x + y)color(red)(cancel(color(black)("moles")))) = 0.335#

#x/(x+y) = 0.335" "" "color(darkorange)((1))#

If you use the molar mass of copper(II) perchlorate and the molar mass of water, you can say that mass of the solution will be equal to the mass of copper(II) perchlorate added to the mass of water.

This sample contains

#x color(red)(cancel(color(black)("moles Cu"("ClO"_4)_2))) * "262.447 g"/(1color(red)(cancel(color(black)("mole Cu"("ClO"_4)_2)))) = (262.447 * x)color(white)(.)"g"#

of copper(II) perchlorate and

#y color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = (18.015 * y)color(white)(.)"g"#

of water, so you can say that

#(262.447 * x) color(red)(cancel(color(black)("g"))) + (18.015 * y)color(red)(cancel(color(black)("g"))) = 1150 color(red)(cancel(color(black)("g")))#

#262.447x + 18.015y = 1150" " " "color(darkorange)((2))#

You now have two equations with two unknowns. Since you're looking for the number of moles of copper(II) perchlorate, use equation #color(darkorange)((1))# to say that

#x = 0.335 * (x + y) implies y = 0.665/0.335 * x#

Plug this into equation #color(darkorange)((2))# to get

#262.447 * x + 18.015 * 0.665/0.335 * x = 1150#

Solve for #x# to find

#298.21 * x= 1150 implies x- 1150/298.21 = 3.856#

Since this represents the number of moles of copper(II0 perchlorate present in #10^3color(white)(.)"mL" = "1 L"# of this solution, you can say that the molarity of the solution is equal to

#color(darkgreen)(ul(color(black)("molarity = 3.86 mol L"^(-1))))#

The answer is rounded to three sig figs.