Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24° C is ________ kg (Molar mass in g mol⁻¹ for ethylene glycol 62, K_f of water =1.86 K kg mol⁻¹ )

Solution

Chemistry ma’am · 4 steps

Turn the 24 degree drop into molality, then into moles of glycol in 18.6 kg of water.

1. Depression needed. Water freezes at 0°C and must now freeze at -24°C, so Δ T_f = 24 K. Ethylene glycol is a non-electrolyte, so i = 1.

   Δ T_f = K_f × nₛₒₗᵤₜₑ/(W_(water) (kg))

2. Moles of glycol. Rearrange: n = (Δ T_f × W)/K_f = (24 × 18.6)/1.86 = 24 × 10 = 240 mol.
3. Convert to mass. Mass = 240 × 62 = 14880 g = 14.88 kg.
4. So. About 14.88 kg of ethylene glycol is needed, which rounds to 15 kg.

The common slip: 14880. Stopping at the mass in grams gives 14880, but the blank asks for kilograms, so divide by 1000 to get about 15.

The idea to keep. For freezing-point depression, molality uses kilograms of solvent. Find moles of solute from Δ T_f = K_f m first, then multiply by molar mass.

Answer from the NTA final answer key. Solution by PrepWiser.