Answer :
Final answer:
The boiling point of a solution made by dissolving 3 moles of KBr in 2000 g of water is calculated as 101.536°C using colligative properties and the boiling point elevation formula. The closest given answer to this calculated value is D. 101.5°C.
Explanation:
The question is asking to determine the boiling point of an aqueous solution containing 3 moles of KBr. To find this, we use colligative properties, in particular, the boiling point elevation formula ΔT = i * Kb * m, where ΔT is the boiling point elevation, i is the van't Hoff factor (which is 2 for KBr, as it dissociates into K+ and Br- ions), Kb is the ebullioscopic constant for the solvent, and m is the molality of the solution.
First, we calculate the molality (m). Molality is defined as moles of solute per kilograms of solvent. In this case,
Molality (m) = moles of KBr / mass of water (kg) = 3 moles / 2 kg = 1.5 m.
Next, we apply the formula to calculate the boiling point elevation:
ΔT = i * Kb * m
= 2 * 0.512 °C/m * 1.5 m
= 1.536 °C.
The normal boiling point of water is 100°C, so the boiling point of the solution is 100°C + 1.536°C = 101.536°C. Therefore, the closest answer from the provided options is 101.5°C.
