The freezing point of pure water is 0.0°C. How many grams of ethylene glycol (C2H6O2) must be mixed in 100.0 g of water to lower the freezing point of the solution to -8.8°C? g

17.54 g

Explanation:

the freezing point of the solution to -8.8°C

depression in freezing point = 8.8°C

Molal freezing point depression constant of ethylene glycol  Kf = 3.11⁰C /m

ΔTf = Kf x m , m is no of moles of solute per kg of solvent .

Let gram of ethylene glycol required be x .

moles = m / mol weight of ethylene glycol

= x / 62

100 g of water = .1 kg

moles of solute dissolved in 1 kg of water

m = x / 62 x .1

m = 10 x / 62

Using the above equation for depression in freezing point

8.8 = 3.11 x 10 x / 62

x = 17.54 g .

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