A solution of sodium chloride in water has a vapor pressure of \(19.6\) torr at \(25^{\circ} \mathrm{C}\). What is the mole fraction of solute particles in this solution? What would be the vapor pressure of this solution at \(45^{\circ} \mathrm{C} ?\) The vapor pressure of pure water is \(23.8\) torr at \(25^{\circ} \mathrm{C}\) and \(71.9\) torr at \(45^{\circ} \mathrm{C}\), and assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.

First, let's find the mole fraction of water (H2O) in the sodium chloride (NaCl) solution. The vapor pressure of the solution is given as 19.6 torr at 25°C. We know the vapor pressure of pure water at 25°C is 23.8 torr. Using Raoult's Law, we can write: \(P_\text{solution} = x_\text{H2O} \times P_\text{H2O}^\text{pure}\) where: \(P_\text{solution}\) = vapor pressure of the solution (19.6 torr), \(x_\text{H2O}\) = mole fraction of water in the solution, \(P_\text{H2O}^\text{pure}\) = vapor pressure of pure water (23.8 torr). Now, we need to solve for the mole fraction of water: \(x_\text{H2O} = \frac{P_\text{solution}}{P_\text{H2O}^\text{pure}}\)

Now that we have the mole fraction of water, we can calculate the mole fraction of sodium chloride (NaCl) in the solution. Since there are only two components in the solution (water and NaCl), their mole fractions should sum up to 1: \(x_\text{NaCl} = 1 - x_\text{H2O}\) However, keep in mind that in the solution, sodium chloride exists as Na+ and Cl- ions. Therefore, we need to multiply the mole fraction of NaCl by 2 to account for both ions: \(x_\text{solute} = 2 \times x_\text{NaCl}\)

Now that we have the mole fraction of solute particles (Na+ and Cl- ions), we can determine the vapor pressure of the solution at 45°C. We know the vapor pressure of pure water at 45°C is 71.9 torr. Using Raoult's Law, we can write: \(P_\text{solution}^\text{45°C} = x_\text{H2O} \times P_\text{H2O}^\text{pure}(\text{45°C})\) where: \(P_\text{solution}^\text{45°C}\) = vapor pressure of the solution at 45°C, \(P_\text{H2O}^\text{pure}(\text{45°C})\) = vapor pressure of pure water at 45°C (71.9 torr). Now, we need to solve the equation to find the vapor pressure of the solution at 45°C: \(P_\text{solution}^\text{45°C} = x_\text{H2O} \times P_\text{H2O}^\text{pure}(\text{45°C})\) Let's plug in the values we calculated in the previous steps to find the final answer.