Find the freezing point of a solution that contains 23.0g ethlene glycol(C2H4(OH)2) dissolved in 250g of ..?
-113
-117
-125
-119
which one
Answer:
The freezing point of pure ethyl ether is
Tf₀ = -116.3°C
The mixture has a lower freezing point. The freezing point depression is given by:
ΔTf = i · Kf · γ · c
i is the van't Hoff factor of the solute. For this solution i=1, because ethylene glycol does dissociate in ethyl ether.
Kf is the cryoscopic constant of the solvent. Kf = -1.79°Cmol/kg for ethyl ether (see2nd link)
γ is the activity coefficient of the solute. γ=1 for ideal mixtures.
A dilute mixture like this one can regarded as ideal.
c is the molality of the solute (= moles solute per kilogram solvent)
The number of of moles of ethylen glycol is:
N = m/M = 23g / 62g/mol = 0.371mol
Its molality in the solution is:
c = 0.371mol / 0.25kg = 1.484mol/kg
The change of the freezing temperature is:
ΔTf = 1 · (-1.79°Cmol/kg) · 1 · 1.484mol/kg = -2.7°C
The freezing point of the mixture is
Tf = Tf₀+ ΔTf = -116.3°C - 2.7°C = -119°C
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: