Final answer:
To calculate the total vapor pressure above a mixture of ethanol and water, we can use Raoult's law. The vapor pressure of ethanol and water can be calculated using their mole fractions and their respective vapor pressures at that temperature. Summing the vapor pressures of ethanol and water will give the total vapor pressure above the mixture.
Step-by-step explanation:
To calculate the total vapor pressure above a mixture of ethanol and water, we need to calculate the mole fraction of each component and use Raoult's law. The mole fraction of ethanol can be calculated by dividing the moles of ethanol by the total moles of the mixture. The moles of ethanol can be calculated by dividing its mass by its molar mass (92.0 g / 46.0 g/mol = 2.0 mol).
The moles of water can be calculated by dividing its mass by its molar mass (186 g / 18.0 g/mol = 10.3 mol). The total moles of the mixture is the sum of the moles of ethanol and water (2.0 mol + 10.3 mol = 12.3 mol). The mole fraction of ethanol is therefore 2.0 mol / 12.3 mol = 0.163. The mole fraction of water is 10.3 mol / 12.3 mol = 0.837.
Raoult's law states that the vapor pressure of a component in a mixture is equal to the mole fraction of the component multiplied by its vapor pressure at that temperature. The vapor pressure of ethanol above the mixture is 0.163 * 144 mm Hg = 23.5 mm Hg.
The vapor pressure of water above the mixture is 0.837 * 118 mm Hg = 98.6 mm Hg. The total vapor pressure above the mixture is the sum of the vapor pressures of the components, which is 23.5 mm Hg + 98.6 mm Hg = 122 mm Hg. Therefore, the correct answer is (d) 122 mm Hg.