To determine the minimum percentage of ethanol needed in the mixture and the pressure at which it should be boiled to achieve a boiling point of 80C, we can use the concept of Raoult's Law. Raoult's Law states that the partial vapor pressure of each component in a solution is equal to the product of the mole fraction of the component and its vapor pressure as a pure substance.Let x1 be the mole fraction of water and x2 be the mole fraction of ethanol in the mixture. Since there are only two components, x1 + x2 = 1.The vapor pressure of pure water P1 at 80C can be found using the Antoine equation, which is given by:log10 P1 = A - B / C + T For water, A = 8.07131, B = 1730.63, and C = 233.426. The temperature, T, should be in Celsius.log10 P1 = 8.07131 - 1730.63 / 233.426 + 80 P1 = 10^2.337 = 217.42 mmHgSimilarly, the vapor pressure of pure ethanol P2 at 80C can be found using the Antoine equation with the constants for ethanol: A = 8.20417, B = 1642.89, and C = 230.3.log10 P2 = 8.20417 - 1642.89 / 230.3 + 80 P2 = 10^2.883 = 760.96 mmHgNow, we can apply Raoult's Law to find the mole fractions of water and ethanol:P_total = x1 * P1 + x2 * P2Since we want the mixture to boil at 80C, the total pressure P_total should be equal to the atmospheric pressure at the boiling point, which is approximately 760 mmHg.760 = x1 * 217.42 + 1 - x1 * 760.96Solving for x1, we get:x1 = 0.914Now, we can find the mole fraction of ethanol x2 :x2 = 1 - x1 = 1 - 0.914 = 0.086To find the percentage of ethanol by volume, we can use the density of each component to calculate the volume fraction. The density of water at 80C is approximately 0.9718 g/mL, and the density of ethanol at 80C is approximately 0.7893 g/mL.Let V1 and V2 be the volume fractions of water and ethanol, respectively. We can write the following equation:x1 = V1 * 0.9718 / V1 * 0.9718 + V2 * 0.7893 Substituting x1 = 0.914, we get:0.914 = V1 * 0.9718 / V1 * 0.9718 + V2 * 0.7893 Solving for V1 and V2, we get:V1 = 0.9718 * V2 / 0.7893 * 1 - 0.914 V1 = 4.64 * V2Now, the volume percentage of ethanol can be calculated as:% ethanol = V2 / V1 + V2 * 100% ethanol = 1 / 4.64 + 1 * 100 = 17.76%Therefore, the minimum percentage of ethanol needed in the mixture is approximately 17.76% by volume. The mixture should be boiled at atmospheric pressure approximately 760 mmHg to achieve a boiling point of 80C.