A solution is prepared by dissolving 215 grams of methanol, CH₃OH, in 1000. grams of water. What is the boiling point of this solution? [The boiling point elevation constant for water is 0.52°C/mole solute in 1000 g of water] A) 93.0°C B) 96.5°C C) 103.5°C D) 107.0°C

Question

Quizplus · Accepted Answer

First, we need to find the molality of the solution:
Molality = moles of solute / mass of solvent in kg
Mass of solute (methanol) = 215 g
Molar mass of methanol (CH3OH) = 32 g/mol + 1 g/mol + 16 g/mol = 33 g/mol
Number of moles of methanol = 215 g / 33 g/mol = 6.52 mol
Mass of solvent (water) = 1000 g = 1 kg
Molality = 6.52 mol / 1 kg = 6.52 m

Next, we can use the boiling point elevation equation to calculate the boiling point elevation:
ΔTb = Kb * molality
ΔTb = 0.52 °C/mole * 6.52 mol = 3.39 °C

The boiling point elevation is the difference between the boiling point of the solution and the boiling point of the pure solvent (water). Since the boiling point of pure water is 100°C, the boiling point of this solution is:
Boiling point = 100°C + 3.39°C = 103.39°C

Since the answer choices are in increments of 0.5°C, the closest answer is C) 103.5°C.

A Solution Is Prepared by Dissolving 215 Grams of Methanol