What mass of ethylene glycol must be added to 5.5 kg of water to lower the freezing point from 0 degrees celsius to -10.0 degrees celsius?

1835.4 g of ethylene glycol

Explanation:

Colligative property of Freezing point depression:

ΔT = Kf . m . i

i = 1 (We assume ethylene glycol as non electrolytic)

ΔT = 10°C (T° freezing pure solvent - T° freezing solution)

(0° - (-10°C)

10°C = 1.86 °C/m . m

5.37 = m

Molal means moles of solute in 1kg of solvent.

In 1 kg we have 5.37 moles of solute

In 5.5 kg we will have (5.5 . 5.37)/1 =  29.6 moles

Molar mass ethylene glycol = 62.07 g/mol

Mol . molar mass = mass ⇒ 29.6 m . 62.07g/m =1835.4 g