Question

What is the freezing point of radiator fluid that is 50% antifreeze by mass? Kf for water is 1.86 ∘C/m.

Answer 1

Answer:

Explanation:

Supposing the antifreeze to be ethylene glycol, C2H6O2.

Take a hypothetical sample of 2000 g of radiator fluid.  Then there is 1000 g each of water and ethylene glycol:

(1000 g C2H6O2) / (62.06810 g C2H6O2/mol) = 16.111 mol C2H6O2

So the molality is 16.111 m.

(1.86 °C/m) x (16.111 m) = 30.0 °C change

0°C - 30.0°C = -30.0°C

(0.512 °C/m) x (16.111 m) = 8.25°C change

100°C + 8.25°C = 108°C

Answer 2

Answer:

Supposing the antifreeze to be ethylene glycol, C2H6O2.

Take a hypothetical sample of 2000 g of radiator fluid. Then there is 1000 g each of water and ethylene glycol:

(1000 g C2H6O2) / (62.06810 g C2H6O2/mol) = 16.111 mol C2H6O2

So the molality is 16.111 m.

(1.86 °C/m) x (16.111 m) = 30.0 °C change

0°C - 30.0°C = -30.0°C

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(0.512 °C/m) x (16.111 m) = 8.25°C change

100°C + 8.25°C = 108°C.