Question

A chemist has 400 grams of salt solution that is 10% salt. How many grams of 20% salt solution must be added to obtain a 12% solution of salt?

Answer 1

You need to add 100 g of the 20% salt solution.

Explanation:

So, you're dealing with two salt solutions of different percent concentrations by mass.

Start by calculating how much salt you get in the 400-g sample of the 10% solution.

`m_"salt"/m_"solution" * 100 = 10%`

`m_"salt" = (10 * m_"solution")/100`

`m_"salt" = (10 * 400)/100 = "40 g salt"`

Now, let's say that the mass of the 20% solution needed is equal to `x` grams. SInce this solution has 20 g of salt for every 100 g of solution, you can say that

`xcolor(red)(cancel(color(black)("g solution"))) * "20 g salt"/(100color(red)(cancel(color(black)("g solution")))) = 20/100x = x/5" g salt"`

The taol mass of the salt in the target 12% solution will be

`m_"salt" = 40 + x/5`

The total mass of the target solution will be

`m_"sol" = 400 + x`

This means that you can write

`((40 + x/5)"g salt")/((400 + x)" g solution") * 100 = 12%`

Rearrange and solve this equation for `x` to get

`(40 + x/5) * 100 = 12 * (400 + x)`

`4000 + 20x = 4800 + 12x`

`8x = 800 => x = 800/8 = color(green)("100 g")`

This means that if you add 100 g of the 20% solution to 400g of the 10% solution, you will get 500 g of a 12% salt solution.