Question

When $ 50ml $ of liquid water at $ {25^o}C $ is added to $ 50ml $ of ethanol (ethyl alcohol), also at $ {25^o}C $ , the combined volume of the mixture is considerably less than 100mL. What is a possible explanation for this $ ? $

Answer 1

Hint: First we know volume is a physical property that depends on the interactions of molecules. Hence volume is not an additive property. Alcohol is soluble in water because alcohol and water are non-additive volumes.

Complete answer:
Since ethanol is an alcohol. Hence ethanol is soluble in water. If we mix $ 50ml $ of water with $ 50ml $ of ethanol at $ {25^o}C $ , then the total volume (obtained) is less than $ 100ml $ (actual). Because the molecules of ethanol are smaller than the molecules of water. So, the ethanol molecules cram themselves between the water molecules. Then the volume is less than expected.
Since the density of water is $ 0.99705 $ $ g/ml $ , the density of ethanol is $ 0.78522 $ $ g/ml $ at $ {25^o}C $
Given, total volume $ {V_{total}} = 100ml $ where the volume of water is $ 50ml $ and the volume of ethanol is $ 50ml $ .
We know that
 $ Density = \dfrac{{Mass}}{{Volume}} $
 $ \Rightarrow Mass = Density \times Volume $ ----(1)
Then using the equation (1)
Mass of water $ = 0.99705 \times 50 = 49.8525g $
Mass of ethanol $ = 0.78522 \times 50 = 39.261g $
Hence the total mass $ = 49.8525 + 39.261 = 89.1135g $
Then, Water $ \% = \dfrac{{49.8525}}{{89.1135}} \times 100 = 55.9427 $ and Ethanol $ \% = \dfrac{{39.261}}{{89.1135}} \times 100 = 44.0573 $
From literature can be found the mixture density at $ 44\% $ wt ( $ {25^o}C $ ) of Ethanol $ = 0.92571\,g/ml $
Hence, the volume of mixture $ = \dfrac{{89.1135}}{{0.92571}} = 96.265\;ml $ .

Note:
The given problem is similar to thinking of a container of sand poured into a container of rocks. The sand fills in the empty spaces between the rocks. As the volume decreases, the density increases for the same mass. As the molecules pack tighter together, the density of the solution becomes more.