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 denote the mole fraction of water in the solution as \(x_{H_2O}\) and the mole fraction of sodium chloride ions (\(\text{Na}^+\) and \(\text{Cl}^-\)) as \(x_{NaCl}\). According to Raoult's law, the vapor pressure of the solution (\(P_{solution}\)) is given by: \[P_{solution} = x_{H_2O} \cdot P^{0}_{H_2O}\] Here, \(P^{0}_{H_2O}\) is the vapor pressure of pure water (23.8 torr at 25°C). We are given \(P_{solution}\) as 19.6 torr at 25°C, and we can use the above equation to calculate the mole fraction of water in the solution: \[x_{H_2O} = \frac{P_{solution}}{P^{0}_{H_2O}} = \frac{19.6}{23.8}\] Now, calculate the value of \(x_{H_2O}\): \(x_{H_2O} = 0.8235\) Since the mole fractions in the solution must sum up to 1, the mole fraction of sodium chloride ions in the solution is given by: \[x_{NaCl} = 1 - x_{H_2O} = 1 - 0.8235\] Calculate the value of \(x_{NaCl}\): \(x_{NaCl} = 0.1765\) Since sodium chloride exists as \(\text{Na}^{+}\) and \(\text{Cl}^{-}\) ions in the solution, the mole fraction of solute particles in the solution is twice the value of \(x_{NaCl}\): \(x_{solute} = 2 \cdot x_{NaCl} = 2 \cdot 0.1765\) Now, calculate the value of \(x_{solute}\): \(x_{solute} = 0.353\) The mole fraction of solute particles in the solution is 0.353.

Now, let's find the vapor pressure of the solution at 45°C. We are given the vapor pressure of pure water at 45°C, which is 71.9 torr. Using Raoult's law again, we can find the vapor pressure of the solution at 45°C: \[P_{solution_{45}} = x_{H_2O} \cdot P^{0}_{H_2O_{45}}\] Here, \(P^{0}_{H_2O_{45}}\) is the vapor pressure of pure water (71.9 torr at 45°C). Now, plug in the values of \(x_{H_2O}\) and \(P^{0}_{H_2O_{45}}\) in the above equation: \[P_{solution_{45}} = 0.8235 \cdot 71.9\] Calculate the value of \(P_{solution_{45}}\): \(P_{solution_{45}} = 59.25\) The vapor pressure of the sodium chloride solution at 45°C is 59.25 torr.