Coolant used in car radiator is aqueous solution of ethylene glycol. In order to prevent the solution from freezing at -0.3° C. How much ethylene glycol must be added to 5 kg of water ? ( Kf =1 ⋅ 86 Kg mol -1)

Question

Coolant used in car radiator is aqueous solution of ethylene glycol. In order to prevent the solution from freezing at -0.3° C. How much ethylene glycol must be added to 5 kg of water ? ( Kf =1 ⋅ 86 Kg mol -1) (A) 50 g (B) 55 g (C) 45 g (D) 40 g

Tardigrade Admin · Accepted Answer

Let W g of ethylene glycol is added. The molar mass of ethylene glycol =62 g / mol. The depression in the freezing point Δ T f = K f m 0.3=1.86 × ( W /62 g / mol × 5 kg ) W =50 g

Q. Coolant used in car radiator is aqueous solution of ethylene glycol. In order to prevent the solution from freezing at $- 0. 3^{\circ} C$ . How much ethylene glycol must be added to $5 k g$ of water ? $(K_{f} = 1 \cdot 86 K g m o l^{- 1})$

50 g

55 g

45 g

40 g

Let $W g$ of ethylene glycol is added.
The molar mass of ethylene glycol $= 62 g / m o l$ .
The depression in the freezing point $Δ T_{f} = K_{f} m$
$0.3 = 1.86 \times \frac{W}{62 g / m o l \times 5 k g}$
$W = 50 g$

Q. Coolant used in car radiator is aqueous solution of ethylene glycol. In order to prevent the solution from freezing at −0.3∘C. How much ethylene glycol must be added to 5kg of water ? (Kf​=1⋅86Kgmol−1)