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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