Question

I'm really bad with story problems, can someone help me out with this one please... A chemist needs 6 liters of a 50% salt solution. All she has available is a 20% salt solution and a 70%
salt solution. How much of each of the two solutions should she mix to obtain her desired solution?

Answer 1

Okay, these problems are easy to solve. You just don't realize that yet :-)

You've got solution A and solution B and you want to make solution C by mixing some of A and B. Here's the trick: figure out exactly how much of the substance is contained in your known volume of solution A, and how much in your known volume of solution B. Solution C's concentration will be [amount of substance in A + amount of substance in B]/[total volume of A + B]

So, here we have solution A, which is 20% salt (0.20), solution B, which is 70% salt (0.70), and we are trying to make 6 liters of solution C, which is 50% salt (0.50)

We know that we want to make 6 liters of solution C, so

A + B = 6

We also know that the amount of salt in A is going to be 0.2 * A, and the salt in B is going to be 0.7 * B, and the salt in C is going to be 0.5 * C = 0.5 * 6 = 3.0. But the salt in C is just the salt in A plus the salt in B:

0.2 * A + 0.7 * B = 3.0

Let's multiply both sides of that equation by 10 to get rid of the pesky decimals:

2A + 7B = 30
A + B = 6

We've got two equations in two unknowns. Solve by elimination or substitution. I would suggest multiplying both sides of the second equation by -2, then adding the two equations together. You'll end up with 1 equation in terms of B only, and you can get the value of B. Then either equation can be used with the known value of B to find A.

Answer 2

1.8 liters of the 20% solution; 4.2 liters of the 70% solution?

Answer 3

Hmm, want to show your work? that's not what I get.