To calculate the partial molar volume of ethanol, first calculate the number of moles of water and ethanol, and determine the volume contributions of both. Afterward, subtract the volume contribution of water from the total volume to find the volume contribution of ethanol, and divide this by the number of moles of ethanol to find its partial molar volume.
The question asks for the calculation of the partial molar volume of ethanol in a 50 percent by mass ethanol-water solution with a known density and partial molar volume of water. First, we need to determine the total mass of the solution and the mass of both ethanol (EtOH) and water.
Given that the density of the solution is 0.914 g/cm3, and we have a 50 percent by mass ethanol-water solution, we can say that 50 g of ethanol and 50 g of water constitute 100 g of the solution. Considering the density, this implies a total volume of about 109.41 cm3 for the 100 g solution (100 g / 0.914 g/cm3).
We are also given the partial molar volume of water is 17.4 cm3 mol-1. Given that the molar mass of water is 18.015 g/mol, we can calculate the number of moles of water in 50 g, which will be approximately 2.776 moles. The total volume contributed by water would then be 2.776 moles * 17.4 cm3 mol-1 = 48.32 cm3.
Next, as per the definition of partial molar volumes, the total volume is the sum of the partial volumes of the components. Thus, the volume contributed by ethanol can be found by subtracting the volume contributed by water from the total volume. This gives us the volume contribution of ethanol as 109.41 cm3 - 48.32 cm3 = 61.09 cm3.
Finally, the partial molar volume of ethanol (VEtOH) can be calculated by dividing the volume contribution of ethanol by the number of moles of ethanol (which is found using its molar mass of 46.07 g/mol for the mass of 50 g).