Automative antifreeze is a 60% weight/weight aqueous solution of ethylene glycol (C2H6O2) then its molality is?

Answer: 24.2 moles/kg

The molar mass of ethylene glycol (C2H6O2) = 2(12.01) + 6(1.01) + 2(16.00) = 62.07 g/mol

Number of moles of ethylene glycol = 60 g / 62.07 g/mol ≈ 0.966 moles

Mass of water = 100 g - 60 g = 40 g

Now, we need to convert the mass of water to kilograms:

Mass of water = 40 g * (1 kg / 1000 g) = 0.04 kg

Now, we can calculate the molality:

Molality (m) = moles of solute / mass of solvent in kg

Molality = 0.966 moles / 0.04 kg ≈ 24.15 moles/kg.

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