Question

How many liters of a 30% alcohol solution must be mixed with 20 liters of a 80% solution to get a 70% solution?

Answer 1

`5` `"L"`

Explanation:

I'll assume the given percentages are those by volume (i.e. `"v/v"`)..

We're asked to find the volume, in `"L"`, of a `30%` `"v/v"` alcohol solution must be added to `20` `"L"` of a `80%` `"v/v"` to obtain a solution that is `70%` `"v/v"`.

To do this, we can set up sort of an algebraic equation that represents this situation.

Here, we have

• `x` liters of `0.3` fractional volume solution

• `20` liters of `0.8` fractional volume solution

These two added together must be a solution with fractional volume `0.7` (the volume will be `20` `"L"` `+ x`), so we can write an equation

`ul(overbrace(x)^"volume of 30%"(0.3) + overbrace((20))^"volume of 80%"(0.8) = overbrace((x+20))^"total volume"(0.7)`

Now, we simply solve for `x`:

`0.3x + 16 = 0.7x + 14`

`0.4x = 2`

`color(red)(ul(x = 5`

So the required volume of the `30%` by volume solution is

`color(red)(ulbar(|stackrel(" ")(" "5color(white)(l)"L"" ")|)`