0.25 mol of a nonvolatile, nonionic solute dissolved in 300 g of water (Kf = 1.86) will lower the freezing point how many degrees?
A) 0.65 °C
B) 2.2 °C
C) 0.14 °C
D) 1.6 °C
E) 3.2 °C

Question

Quizplus · Accepted Answer

Use the formula: ∆Tf = Kf x molality
molality = moles of solute / kg of solvent
Here, moles of solute = 0.25 mol and mass of solvent = 300 g = 0.3 kg
molality = 0.25 mol / 0.3 kg = 0.833 mol/kg
∆Tf = 1.86 x 0.833 = 1.55 °C (rounded to two decimal places)
Therefore, the freezing point will be lowered by 1.6 °C (rounded to one decimal place). The correct answer is D.

025 Mol of a Nonvolatile, Nonionic Solute Dissolved in 300