Question

# How many grams of ethylene glycol (C2H6O2) must be added to 1.10 kg of water to...

How many grams of ethylene glycol (C2H6O2) must be added to 1.10 kg of water to produce a solution that freezes at -5.24 ∘C?

Ans :- Grams of ethylene glycol required = 192.13 g

Explanation :-

From Depression in freezing point , we have

ΔTf = i x kf x m ..................(1)

here, ΔTf = Depression in freezing point = Tf0 - Tf

Tf0 = freezing point of pure solvent = 00C and

Tf = freezing point of solution = - 5.24 0C

kf = Molal freezing point constant = 1.86°C mol-1 kg for H2O

i= Vant's haff factor = 1 (for non-electrolyte like ethylene glycol )

m = Molality = Number of moles / Mass of solvent in kg

Number of moles of ethylene glycol = Mass (W) of ethylene glycol / Gram molar mass of ethylene glycol and

Gram molar mass of ethylene glycol = 62 g/mol

Substitute all these values in equation (1), We have

0 - (-5.24 0C) = 1 x 1.86°C mol-1 Kg x W / 62 g mol-1 x 1.10 Kg

5.24 = W x 0.027272727 g-1

W = 5.24 / 0.027272727 g-1

W = 5.24 / 0.027272727 g-1

W = 192.13 g

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