# How much salt must be added to 800 grams of a solution that is 15% to obtain a solution that is 20% salt?

Aug 24, 2015

You would add 50 g of salt.

#### Explanation:

We will assume that the density of the solution remains constant (even though the density changes with concentration!).

A 15 % concentration means

15 % = "15 g salt"/"100 g solution"

So the mass of salt in 800 g of solution is

$\text{Mass of salt" = 800 color(red)(cancel(color(black)("g soln"))) × "15 g salt"/(100 color(red)(cancel(color(black)("g soln")))) = "120 g salt}$

So you have $\text{120 g salt"/"800 g soln}$.

You want to add $x \text{ g of salt}$ to get a 20 % solution.

This also increases the mass of the solution by $x \text{ g}$.

Then

$\frac{120 + x}{800 + x} = \frac{20}{100}$

$100 \left(120 + x\right) = 20 \left(800 + x\right)$

$12000 + 100 x = 16000 + 20 x$

$80 x = 5000$

$x = 50$

You would have to add 50 g of salt.

Check:

$\frac{120 + 50}{800 + 50} = \frac{20}{100}$

$\frac{170}{850} = \frac{20}{100}$

$\frac{1}{5} = \frac{1}{5}$

It checks!