A solution contains 25% copper sulphate. Another solution contains 50% copper sulphate. How much of each solution should be used to make 1000 mL of a solution that's 45% copper sulphate?

1 Answer
Aug 9, 2015

Answer:

You need 600 mL of 25% solution and 600 mL of 50% solution.

Explanation:

You need to use the information given to you to write two equations, one that focuses on the mass of the individual solutions that you mix, and the other that focuses on how much copper sulfate the target solution must contain.

Let's say that #x# represetns the volume of the 25% solution and #y# the volume of the 50% solution. The mass of the final solution is 1000 mL, so you can write

#x + y = 1000#

Now, the target solution is known to 45% copper sulfate. This means that the amount of copper sulfate you get in 1000 mL is equal to

#1000color(red)(cancel(color(black)("mL solution"))) * "45 g copper sulfate"/(100color(red)(cancel(color(black)("mL solution")))) = "450 g"# #CuSO_4#

The amount of copper sulfate you get from each solution must amount to 450 g. This means that you can write

#25/100 * x + 50/100 * y = 450#

The two equations are

#{(x + y = 1000), (1/4x + 1/2y = 450) :}#

Use the first equation to write #x# as a function of #y#, then the second equation to solve for #y#

#x = 1000 - y#

#1/4 * (1000-y) + 1/2y = 450#

#250 - 1/4y + 1/2y = 450#

#1/2y = 200 implies y = 200 * 2 = color(green)(400)#

This means that #x# is equal to

#x = 1000 - 400 = color(green)(600)#

Therefore, if you mix 600 mL of 25% solution with 400 mL of 50% solution you get a solution that is 45% copper sulfate.