answer: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 80°C, 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 80°C 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 mmHg Similarly, the vapor pressure of pure ethanol (P2) at 80°C 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 mmHg Now, we can apply Raoult's Law to find the mole fractions of water and ethanol: P_total = x1 * P1 + x2 * P2 Since we want the mixture to boil at 80°C, 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) * 760.96 Solving for x1, we get: x1 = 0.914 Now, we can find the mole fraction of ethanol (x2): x2 = 1 - x1 = 1 - 0.914 = 0.086 To 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 80°C is approximately 0.9718 g/mL, and the density of ethanol at 80°C 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 * V2 Now, 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 80°C.

answer:To solve this problem, we will use the concept of phase diagrams and the lever rule. We will first determine the melting points of pure benzene and toluene, and then use these values to estimate the melting point of the mixture. We will also determine the composition of the solid and liquid phases at equilibrium. Step 1: Determine the melting points of pure benzene and toluene The melting point of pure benzene is 5.5°C, and the melting point of pure toluene is -95°C. Step 2: Estimate the melting point of the mixture Since we have a 50 mol% benzene and 50 mol% toluene mixture, we can use a linear interpolation to estimate the melting point of the mixture. This is based on the assumption that the melting point of the mixture varies linearly with the composition. Melting point of the mixture = (0.5 * 5.5) + (0.5 * -95) = 2.75 - 47.5 = -44.75°C Step 3: Determine the composition of the solid and liquid phases at equilibrium To determine the composition of the solid and liquid phases at equilibrium, we will use the lever rule. The lever rule states that the ratio of the lengths of the tie lines in a phase diagram is equal to the ratio of the amounts of the two phases present at equilibrium. Let xB and xT be the mole fractions of benzene and toluene in the solid phase, and yB and yT be the mole fractions of benzene and toluene in the liquid phase. At equilibrium, the chemical potential of benzene and toluene in the solid phase must be equal to their chemical potential in the liquid phase. μB(solid) = μB(liquid) μT(solid) = μT(liquid) We can use the Gibbs-Duhem equation to relate the chemical potentials in the solid and liquid phases: xB * dμB(solid) + xT * dμT(solid) = 0 yB * dμB(liquid) + yT * dμT(liquid) = 0 Since the chemical potentials of benzene and toluene are equal in both phases, we can equate the two equations: xB * dμB(solid) + xT * dμT(solid) = yB * dμB(liquid) + yT * dμT(liquid) We can rearrange this equation to solve for the mole fractions of benzene and toluene in the solid and liquid phases: xB = yB * (1 - xT) / xT xT = yT * (1 - yB) / yB Since we know the overall composition of the mixture (50 mol% benzene and 50 mol% toluene), we can use this information to solve for the mole fractions of benzene and toluene in the solid and liquid phases: xB + xT = 0.5 yB + yT = 0.5 Solving these equations simultaneously, we get: xB = 0.25 xT = 0.25 yB = 0.25 yT = 0.25 Therefore, the composition of the solid phase at equilibrium is 25 mol% benzene and 25 mol% toluene, and the composition of the liquid phase at equilibrium is also 25 mol% benzene and 25 mol% toluene. In conclusion, the melting point of a mixture of 50 mol% benzene and 50 mol% toluene at a pressure of 1 atm is approximately -44.75°C. The composition of the solid and liquid phases at equilibrium is 25 mol% benzene and 25 mol% toluene for both phases.

answer:The effect of pressure on the melting point of water can be explained using the phase diagram of water and the Clausius-Clapeyron equation. In general, increasing pressure leads to a slight increase in the melting point of water, but the effect is not as significant as the effect of pressure on the boiling point of water. The phase diagram of water shows the relationship between pressure and temperature for different phases of water (solid, liquid, and gas). The line separating the solid and liquid phases is called the melting curve. The slope of this curve indicates how the melting point of water changes with pressure. The Clausius-Clapeyron equation relates the change in pressure (ΔP) to the change in temperature (ΔT) for a phase transition, such as melting: ΔP/ΔT = L/(T*ΔV) where L is the latent heat of the phase transition, T is the temperature, and ΔV is the change in volume during the phase transition. For the melting of ice to water, the change in volume (ΔV) is negative, as ice has a lower density than water. This means that the slope of the melting curve is positive, indicating that the melting point of water increases with increasing pressure. However, the magnitude of this effect is relatively small because the change in volume during melting is small compared to the change in volume during vaporization. As pressure increases, water molecules in the ice lattice are forced closer together, which makes it more difficult for them to break free from the lattice and transition into the liquid phase. This is why the melting point of water increases with pressure. However, the effect is not linear, and the relationship between pressure and melting point is more complex than a simple linear relationship. In summary, the effect of pressure on the melting point of water is that it increases the melting point, but the effect is relatively small compared to the effect of pressure on the boiling point of water. This is due to the small change in volume during the melting process and the complex relationship between pressure and melting point.

answer:To determine the pressure at which water boils at a temperature of 98 °C, we can use the Antoine equation, which is a semi-empirical equation that relates the vapor pressure of a substance to its temperature. For water, the Antoine equation constants are: A = 8.07131 B = 1730.63 C = 233.426 The Antoine equation is given by: log10(P) = A - (B / (T + C)) where P is the vapor pressure in mmHg and T is the temperature in °C. Plugging in the given temperature (T = 98 °C) into the equation: log10(P) = 8.07131 - (1730.63 / (98 + 233.426)) log10(P) = 8.07131 - (1730.63 / 331.426) log10(P) = 8.07131 - 5.22097 log10(P) = 2.85034 Now, we need to find the pressure P: P = 10^(2.85034) P ≈ 717.68 mmHg To convert this pressure to atmospheres (atm), we can use the conversion factor: 1 atm = 760 mmHg P(atm) = 717.68 mmHg * (1 atm / 760 mmHg) P(atm) ≈ 0.9443 atm So, the pressure at which water boils at a temperature of 98 °C is approximately 0.9443 atm.