What is the value of the van’t Hoff factor for KCl if a 1.00 m aqueous solution shows a vapor pressure depression of 0.734 mm Hg at 298 K? (The vapor pressure of water at 298 K is 23.76 mm Hg.)

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1. Identify the formula for vapor pressure depression: \( \Delta P = i \cdot m \cdot K_f \), where \( \Delta P \) is the vapor pressure depression, \( i \) is the van’t Hoff factor, \( m \) is the molality, and \( K_f \) is the cryoscopic constant. However, since we are dealing with vapor pressure, we use \( \Delta P = i \cdot m \cdot P^0 \), where \( P^0 \) is the vapor pressure of the pure solvent.
2. Calculate the vapor pressure depression \( \Delta P \) using the given values: \( \Delta P = P^0 - P_{solution} = 23.76 \text{ mm Hg} - (23.76 \text{ mm Hg} - 0.734 \text{ mm Hg}) \).
3. Substitute the known values into the vapor pressure depression formula: \( 0.734 \text{ mm Hg} = i \cdot 1.00 \text{ m} \cdot 23.76 \text{ mm Hg} \).
4. Solve for the van’t Hoff factor \( i \) by rearranging the equation: \( i = \frac{0.734 \text{ mm Hg}}{1.00 \text{ m} \cdot 23.76 \text{ mm Hg}} \).
5. Interpret the result: The van’t Hoff factor \( i \) represents the number of particles the solute dissociates into in solution. For KCl, which dissociates into K\(^+\) and Cl\(^-\), the theoretical value of \( i \) should be close to 2.