What is the change in the boiling point of a solution made by dissolving 14.7 g of C₆H₁₂O₆ into 150.0 mL of water? The density of water is 1.00 g/mL and K_b = 0.512°C/m.

Question

Quizplus · Accepted Answer

First, we need to calculate the molality of the solution:

molality = moles of solute / mass of solvent in kg

mass of solvent = volume of water x density of water = 150.0 mL x 1.00 g/mL = 150.0 g = 0.150 kg

moles of solute = mass of solute / molar mass = 14.7 g / 152.3 g/mol = 0.0965 mol

molality = 0.0965 mol / 0.150 kg = 0.6433 mol/kg

Next, we can use the equation:

ΔTb = Kb x molality

where ΔTb is the change in boiling point, Kb is the boiling point elevation constant (given as 0.512°C/m), and molality is the molality of the solution we just calculated.

ΔTb = 0.512°C/m x 0.6433 mol/kg = 0.3296°C

Therefore, the change in boiling point is 0.279°C, which is answer choice D.

What Is the Change in the Boiling Point of a Solution