Answer to Question #185641 in General Chemistry for Joshua

Question

Congratulations! You won the lotto and bought a new car. Since it's very hot in the philippines you are required to buy an antifreeze to keep the water in your car's radiator from boiling. You add 1.00kg of ethylene glycol (C2H6O2) antifreeze to 4450 g of water in your car's radiator. What are the boiling and freezing points of the solution?

Answer 1

Mass of ethylene glycol= 4000g

Mass of water = 4450g= 4.45Kg

Moles of ethylene glycol= mass/molar mass = 4000g/62g/mol = 64.5mol

Molality= moles of solute/Kg of solvent = 64.5/4.45 = 14.5Molal

Boiling point elevation constant of ethylene glycol = -3.11°C/mol/Kg

Freezing point depression constant of ethylene glycol= -12.9°C/mol/Kg

Boiling point elevation of solution ∆T_b= K_bm

∆T_b= -3.11x14.5= -45.1°C

New boiling point= 100°C -(-45.1°C)=145.1°C

Freezing point depression ∆T_f=K_fm

∆T_f= -12.9×14.5= -187.05

New freezing point = 0°C - 187.05= -187.05°C