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.

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