Question

If you had 0.956 moles of a nonelectrolyte solute in 345 grams of water. What is the new freezing point? (The Kf of water is 1.86 °C/m ) *

Answer 1

solution molarity

$M=\frac{n}{V}=\frac{0.956}{0.345}=2.77$

freezing point depression

$\Delta Tf=Kf\times M=1.86\times2.77=5.15$

freezing point

$\Delta Tfsol=0-5.15=-5.15$ °C