If 200g of ethylene glycol is added to 1000g of water in a car radiator what will be the freezing point of the

Question

solution? Please help

Answer 1

200 g / 62.07 g per mole = 3.22 mole
3.22 molal x -1.86 = -6.0 deg C (21.2 deg F)

Answer 2

C2H4(OH)2 = C2H6O2 molecular mass = 2(12) + 6(1) + 2(16) = 62 grams/mole

Call Td delta T since we can't type Greek here


Td = Kfp X molality = (1.86 C kg / mole)(200 g/ 62 g/mole 1 kg) = 6 C

So the water will freeze at -6 degrees Celsius