What is the van’t Hoff factor for K2SO4 in an aqueous solution that is 5.00% K2SO4 by mass and freezes at -1.21 °C?

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1. Step 1: Understand the van't Hoff factor (i), which represents the number of particles a compound dissociates into in solution. For K2SO4, it dissociates into 2 K+ ions and 1 SO4^2- ion, so theoretically, i = 3.
2. Step 2: Use the freezing point depression formula: \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the change in freezing point, \( K_f \) is the cryoscopic constant of water (1.86 °C kg/mol), and \( m \) is the molality of the solution.
3. Step 3: Calculate the change in freezing point: \( \Delta T_f = 0 - (-1.21) = 1.21 \) °C.
4. Step 4: Determine the molality (m) of the solution. First, calculate the mass of K2SO4 in 100 g of solution (5.00 g), then convert this mass to moles using the molar mass of K2SO4 (174.26 g/mol). Finally, calculate molality by dividing moles of solute by the mass of solvent in kg (95 g of water = 0.095 kg).
5. Step 5: Rearrange the freezing point depression formula to solve for the van't Hoff factor (i): \( i = \frac{\Delta T_f}{K_f \cdot m} \). Substitute the values for \( \Delta T_f \), \( K_f \), and \( m \) to find the experimental van't Hoff factor.