To solve this problem, we can use Raoult's Law, which states that the partial pressure of a component in a mixture is equal to the mole fraction of that component multiplied by its vapor pressure as a pure substance.Let's denote the mole fraction of ethanol as x_ethanol and the mole fraction of water as x_water. Since there are only two components in the mixture, we know that:x_ethanol + x_water = 1We are given the total pressure P_total of the mixture as 760 torr and the vapor pressure P_vapor of the mixture as 45 torr. We can now use Raoult's Law to find the mole fraction of ethanol in the mixture.P_vapor = x_ethanol * P_ethanol + x_water * P_waterWe need to find the vapor pressures of pure ethanol P_ethanol and pure water P_water at the given temperature. These values can be found in reference tables or through experiments. For the sake of this problem, let's assume the vapor pressure of pure ethanol is 100 torr and the vapor pressure of pure water is 50 torr.Now we can plug these values into the equation:45 torr = x_ethanol * 100 torr + x_water * 50 torrSince x_ethanol + x_water = 1, we can rewrite x_water as 1 - x_ethanol :45 torr = x_ethanol * 100 torr + 1 - x_ethanol * 50 torrNow we can solve for x_ethanol:45 torr = 100x_ethanol + 50 - 50x_ethanol45 torr - 50 torr = 100 - 50 x_ethanol-5 torr = 50x_ethanolx_ethanol = -5 torr / 50 = -0.1However, a negative mole fraction is not physically possible. This indicates that there might be an error in the given vapor pressures of pure ethanol and pure water or the vapor pressure of the mixture. Please double-check the given values and try again. If the values are correct, it may indicate that the mixture is not following Raoult's Law, and a different approach may be needed to solve the problem.