Kf 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 (C2H6O2) must you add to get the freezing point of the solution lowered to -2.8°C ?

Question

Kf 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 (C2H6O2) must you add to get the freezing point of the solution lowered to -2.8°C ? (A) 72 g (B) 93 g (C) 39 g (D) 27 g

Tardigrade Admin · Accepted Answer

Δ Tf = i Ã kf Ã m 2.8 = 1 Ã 1.86 Ã (x/62 ×1) x = (2.8 × 62/1.86) = 93 gm

Q. $K_{f}$ for water is $1.86 K k g m o l^{- 1}$ . If your automobile radiator holds $1.0 k g$ 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 $- 2. 8^{\circ} C$ ?

$72 g$

$93 g$

$39 g$

$27 g$

$Δ T_{f} = i \times k_{f} \times m$
$2.8 = 1 \times 1.86 \times \frac{x}{62 \times 1}$
$x = \frac{2.8 \times 62}{1.86} = 93 g m$

Q. Kf​ for water is 1.86Kkgmol−1. If your automobile radiator holds 1.0kg of water, how may grams of ethylene glycol (C2​H6​O2​) must you add to get the freezing point of the solution lowered to −2.8∘C ?

72g

93g

39g

27g