Question

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

Answer 1

To a first approximation, the volumes would indeed add up to `100*mL`....

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 `Mg(OCH_2CH_3)_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_2O)=1.0*g*mL^-1`, `rho_"ethanol"=0.789*g*mL^-1`, and so (finally) we have the quotient.....

`rho=(1.0*g*mL^-1xx50*mL+0.789*g*mL^-1xx50*mL)/(100*mL)`

`0.895*g*mL^-1`. And really the only way could verify this is experimentally, and you would have difficulty in preparing anhydrous ethanol.