Answer:
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
