Question

1.) A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 �C/m? Show all of the work needed to solve this problem.

2.) The freezing point of a solution will ____________ and the boiling point will ____________ with _____________ molality of a nonvolatile solute.

increase, increase, increasing

increase, increase, decreasing

decrease, increase, increasing

decrease, decrease, decreasing

Answer 1

Number of mole=Mass(g)/Molar Mass(g/mol)
25.5/(12*6)+(1*12)+(16*6)=0.1416667 moles
1 grams represent 1 cm^3
398 grams represent 398 cm^3
Number of Moles=Concentration(mol/dm^3)*Volume(dm^3)
Change 398 unit from cm^3 to dm^3 by dividing by 1000
0.141667/0.398=Concentration
ΔT = -Kfm
ΔT = Change in temperature in °C
Kf = freezing point depression constant or cryoscopic constant in °C kg/mol
m = molarity of the solute in mol solute/kg solvent.