# Question #ddc63

Dec 3, 2017

This is an ill-posed problem. Moreover, there aren't enough data to solve any possible specification of it.

#### Explanation:

Let's start from the (very limited) information we have.

1. The volume of a mixture is not the sum of separate components' volumes. Separate volumes (60 mL of water and 40 mL of alcohol) are not knowable ways to measure the amounts of components once you have mixed the liquids. In other words there is no way to ascertain which volume is water and which is alcohol in the solution.
Therefore,
a) the solution can't "have" such volumes and
b) we don't know neither the total volume of solution, nor its mass.

2. We could restate the question, saying that 60 mL of water and 40 mL of alcohol are the volumes of the two pure liquids before mixing. Even in this case we can't solve the problem in absence of another experimental data related to the mass or volume of the resulting solution (such as the solution volume and/or density).

3. Assuming we know the volume of solution (${V}_{\text{mix}}$) we could determine just one kind of percent of water: volume by volume:
$\frac{60}{V} _ \text{mix} \cdot 100$
To calculate other kinds of percent (mass out of mass or mass out of volume) you should know also the densities of pure components.