K_f for water is 1.86 K kg mol^–1. If your automobile radiator holds 1.0 kg of water, how may grams of ethylene glycol (C_2H_6O_2) must you add to get the freezing point of the solution lowered to –2…

Question

K_f for water is 1.86 K kg mol^–1. If your automobile radiator holds 1.0 kg of water, how may grams of ethylene glycol (C_2H_6O_2) must you add to get the freezing point of the solution lowered to –2.8ºC?

Accepted Answer

Correct answer: B. (b) ∆ T_f = i × k_f × m 2.8 = 1 × 1.86 × x62 1x = 2.8 621.86= 93 g

$K_{f} for water is 1.86 K kg m o l^{-1} . If your automobile radiator holds 1.0 kg of water, how may grams of ethylene glycol (C_{2} H_{6} O_{2}) must you add to get the freezing point of the solution lowered to -2.8ºC?$

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