A solution contains \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) and an unknown number of moles of sodium chloride. The vapor pressure of the solution at \(29^{\circ} \mathrm{C}\) is \(3.85 \mathrm{kPa}\). The vapor pressure of pure water at this temperature is \(4.05 \mathrm{kPa}\). Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)

Raoult's Law states that the vapor pressure of a component in an ideal solution is proportional to the mole fraction of that component in the solution. Mathematically, this can be represented as: \[P_A = X_A \times P^*_A\] where: \(P_A\) = vapor pressure of component A in the solution \(X_A\) = mole fraction of component A in the solution \(P^*_A\) = vapor pressure of pure component A We can rearrange this equation to solve for the mole fraction of water: \[X_{H_2O} = \frac{P_{H_2O}}{P^*_{H_2O}}\] Given the vapor pressure of the solution, \(P_{H_2O} = 3.85\, kPa\), and the vapor pressure of pure water, \(P^*_{H_2O} = 4.05 \, kPa\), we can find the mole fraction of water: \[X_{H_2O} = \frac{3.85}{4.05} = 0.9506\]

Since there are only two components in the solution, water and sodium chloride, the sum of their mole fractions must be equal to 1. Therefore, the mole fraction of sodium chloride, \(X_{NaCl}\), can be calculated as: \[X_{NaCl} = 1 - X_{H_2O} = 1 - 0.9506 = 0.0494\]

Sodium chloride is a strong electrolyte, which means it completely dissociates into its constituent ions in the solution. This results in two moles of ions (Na+ and Cl-) for each mole of sodium chloride. Therefore, the mole fraction of solute particles, \(X_s\), is twice the mole fraction of sodium chloride: \[X_s = 2 \times X_{NaCl} = 2 \times 0.0494 = 0.0988\]

Since the moles of water in the solution is given as 0.5 mol, we can use the mole fraction of solute particles to determine the moles of sodium chloride: \[\text{moles of NaCl} = \frac{X_s}{2} \times \text{moles of }\, H_2O\] \[\text{moles of NaCl} = \frac{0.0988}{2} \times 0.5 = 0.0247\, \text{mol}\]

We can now convert the number of moles of sodium chloride to grams using its molecular weight (MW = 58.44 g/mol): \[\text{grams of NaCl} = \text{moles of NaCl} \times \text{molecular weight of NaCl}\] \[\text{grams of NaCl} = 0.0247\, \text{mol} \times 58.44\, \frac{\text{g}}{\text{mol}} = 1.44\, \text{g}\] So there are 1.44 grams of sodium chloride in the solution.