# If 50*mL ethanol were added to 50*mL water, what would the volume and density be of the final solution?

Sep 16, 2017

To a first approximation, the volumes would indeed add up to $100 \cdot m L$....

#### Explanation:

As you know ethanol is a water like solvent, and infinitely miscible in water. And note that the ethanol we typically use is approx. 5-10% water..... Ethanol is exceptionally difficult to dry, and when you DO dry it, typically by distillation from $M g {\left(O C {H}_{2} C {H}_{3}\right)}_{2}$ under a protective atmosphere of argon or dinitrogen, I fancy that when you do expose the ethanol to air, the volume of the liquid increases, as it sucks up atmospheric water.

And so I am quite justified in taking the quotient....

rho_"mixture"="Mass of solution"/("Volume of solution"), and while we know ${\rho}_{{H}_{2} O} = 1.0 \cdot g \cdot m {L}^{-} 1$, ${\rho}_{\text{ethanol}} = 0.789 \cdot g \cdot m {L}^{-} 1$, and so (finally) we have the quotient.....

$\rho = \frac{1.0 \cdot g \cdot m {L}^{-} 1 \times 50 \cdot m L + 0.789 \cdot g \cdot m {L}^{-} 1 \times 50 \cdot m L}{100 \cdot m L}$

$0.895 \cdot g \cdot m {L}^{-} 1$. And really the only way could verify this is experimentally, and you would have difficulty in preparing anhydrous ethanol.