Let's break down the problem step by step as requested. --- ## **Step 1: Write the Van't Hoff Factor Equation** The problem asks for the vapor pressure lowering, ΔP, when .75 × 10⁻² mol of urea is added to .500 kg of water. We use **Raoult's Law** for vapor pressure lowering: \[ \Delta P = X_{\text{solute}} \cdot P^\circ_{\text{solvent}} \] Where: - \( X_{\text{solute}} \) = mole fraction of solute (urea) - \( P^\circ_{\text{solvent}} \) = vapor pressure of pure solvent (water) **Explanation:** Raoult's Law relates the vapor pressure lowering to the mole fraction of a nonvolatile solute. --- ## **Step 2: Calculate Values and Solve** ### **2a. Calculate Moles of Solvent (Water)** - Mass of water = .500 kg = 500 g - Molar mass of water = 18.02 g/mol \[ \text{Moles of water} = \frac{500\, \text{g}}{18.02\, \text{g/mol}} \approx 27.75\, \text{mol} \] ### **2b. Calculate Mole Fraction of Urea** - Moles of urea = 1.75 × 10⁻² mol \[ X_{\text{urea}} = \frac{\text{moles of urea}}{\text{moles of urea} + \text{moles of water}} = \frac{1.75 \times 10^{-2}}{1.75 \times 10^{-2} + 27.75} \] \[ X_{\text{urea}} \approx \frac{.0175}{27.7675} \approx 6.30 \times 10^{-4} \] **Explanation:** The mole fraction is found by dividing the moles of solute by the total moles in the solution. ### **2c. Calculate ΔP** Given: \( P^\circ_{\text{water}} = 23.8 \) torr \[ \Delta P = X_{\text{urea}} \times P^\circ_{\text{water}} = 6.30 \times 10^{-4} \times 23.8 \approx .015 \text{ torr} \] --- ## **Final Answer** \[ \boxed{.015\ \text{torr}} \] **Summary:** - **Step 1:** Write Raoult's Law for vapor pressure lowering. - **Step 2:** Calculate moles, mole fraction, and then use Raoult's Law to find ΔP. - **Final Answer:** The vapor pressure lowering is **.015 torr**.