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

1 Answer
anor277
Sep 16, 2017

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.

Related questions
See all questions in Density
Impact of this question
3293 views around the world
You can reuse this answer
Creative Commons License