Another example of the apparent molar volume of the second component is less than its molar volume as a pure substance is the case of ethanol in water. For example, at 20 mass percents ethanol, the solution has a volume of 1.0326 liters per kg at 20 °C, while pure water is 1.0018 L/kg (1.0018 cc/g). The apparent volume of the added ethanol is 1.0326 L – 0.8 kg x 1.0018 L/kg = 0.2317 L. The number of moles of ethanol is 0.2 kg / (0.04607 kg/mol) = 4.341 mol, so that the apparent molar volume is 0.2317 L / 4.341 mol = 0.0532 L / mol = 53.2 cc/mole (1.16 cc/g). However pure ethanol has a molar volume at this temperature of 58.4 cc/mole (1.27 cc/g).
If the solution were ideal, its volume would be the sum of the unmixed components. The volume of 0.2 kg pure ethanol is 0.2 kg x 1.27 L/kg = 0.254 L, and the volume of 0.8 kg pure water is 0.8 kg x 1.0018 L/kg = 0.80144 L, so the ideal solution volume would be 0.254 L + 0.80144 L = 1.055 L. The nonideality of the solution is reflected by a slight decrease (roughly 2.2%, 1.0326 rather than 1.055 L/kg) in the volume of the combined system upon mixing. As the percent ethanol goes up toward 100%, the apparent molar volume rises to the molar volume of pure ethanol.