K_f for water is 1.86 K kg mol^-1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C₂H₆O₂) must you add to get the freezing point of the solution lowered to -2.8^o C ?

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72 g

27 g

93 g

39 g

answer is B.

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Detailed Solution

Coolant glycol (C₂H₆O₂)is a non-electrolyte.
$∆ T_{f} = 2 . 8^{o} ∆ T_{f} = \frac{100 K_{f} w_{1}}{m_{1} w_{2}} \Rightarrow 2.8 = \frac{1000 \times 1.86 \times w_{1}}{62 \times 1000} ∴ w_{1} = 93.33 g \approx 93 g$

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