Which solution has the same boiling point as 0.25 mol CaCl2 dissolved in 1000 g water?

To find a solution with the same boiling point elevation as 0.25 mol CaCl₂ dissolved in 1000 g of water, we first need to calculate the van 't Hoff factor (i) for CaCl₂, which is 3 (since it dissociates into three ions: one Ca²⁺ and two Cl⁻). The boiling point elevation is determined by the formula ΔT_b = i * K_b * m, where m is the molality. Therefore, a solution with an equivalent molality that produces the same total number of particles (0.75 mol of ions) will have the same boiling point; one such example could be 0.75 mol of NaCl in 1000 g of water, as it also dissociates into two ions (Na⁺ and Cl⁻).