Equal volumes of ethanol (C2H5OH) and water are mixed. If the density of water is 1.00 g·cm–3 and that of ethanol is 0.789, what is the mole fraction of ethanol in the mixture?
Explanation: Let's assume 1L of ethanol and 1L of water since we are told equal volumes are mixed. 1L=1000ml
1ml = 1cm^3
Density = mass /volume
We can use the volumes to find mass of each.
Water: 1.00 g·cm–3= mass/1000ml
mass of water = 1000g
Ethanol: 0.789g·cm–3 = mass/1000ml
mass of ethanol = 789g
Mole fraction of ethanol= Xethanol = moles of ethanol/ total moles
Let's find moles of each by dividing by the molar mass
Moles of water= 1000g/(18g/mol) = 55.56mol
Moles of ethanol = 789g/ (46.07 g/mol) = 17.13mol
Mole fraction of ethanol = 17.13mol/ ( 55.56mol+17.13mol) = 0.236