# 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?

Aug 5, 2015

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

${m}_{\text{salt" = (10 * 400)/100 = "40 g salt}}$

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

${m}_{\text{salt}} = 40 + \frac{x}{5}$

The total mass of the target solution will be

${m}_{\text{sol}} = 400 + x$

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

$\left(40 + \frac{x}{5}\right) \cdot 100 = 12 \cdot \left(400 + x\right)$

$4000 + 20 x = 4800 + 12 x$

$8 x = 800 \implies x = \frac{800}{8} = \textcolor{g r e e n}{\text{100 g}}$

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.