# Question #61472

##### 1 Answer

#### Answer:

#### Explanation:

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

To make the calculations easier, pick a **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

If you take

#(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

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

Plug this into equation

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

Solve for

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

Since this represents the number of moles of copper(II0 perchlorate present in **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**.