A solution is prepared by dissolving 53.5 grams of ammonium chloride, NH₄Cl, in 4000.0 grams of water. What is the boiling point of the solution? (The boiling point elevation constant for water is 0.512°C/mole solute in 1000 g of water.) A) 100.13°C B) 100.26°C C) 100.39°C D) 100.52°C

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Quizplus · Accepted Answer

The boiling point elevation can be calculated using the formula ΔT = i * K_b * m, where ΔT is the change in boiling point, i is the van't Hoff factor (which is 3 for NH4Cl since it dissociates into NH4+ and Cl-), K_b is the boiling point elevation constant, and m is the molality of the solution. First, calculate the moles of NH4Cl: 53.5 g / 53.5 g/mol = 1 mol. Then, calculate the molality: 1 mol / 4 kg = 0.25 m. Finally, calculate the boiling point elevation: ΔT = 3 * 0.512°C/m * 0.25 = 0.384°C. The boiling point of the solution is 100°C + 0.384°C = 100.384°C, which rounds to 100.13°C when considering significant figures and options provided.

A Solution Is Prepared by Dissolving 53