0.25 mol of a nonvolatile,nonionic solute dissolved in 300 g of water (K_f = 1.86 °C m^-1)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

ΔTf = Kf × molality 
where Kf is the freezing point depression constant and is equal to 1.86 °C/m for water
molality (m) = moles of solute / mass of solvent (in kg)

Given, moles of solute = 0.25 mol 
mass of solvent = 300 g = 0.3 kg

Therefore, 
molality (m) = 0.25 mol / 0.3 kg = 0.833 mol/kg

ΔTf = 1.86 °C/m × 0.833 mol/kg = 1.55 °C

So, the freezing point will be lowered by 1.6 °C (rounded to one decimal place). Therefore, the correct answer is D.

025 Mol of a Nonvolatile,nonionic Solute Dissolved in 300 G