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What mass of ethylene glycol (C2H6O2) is required to lower the freezing point of 50.0 g...

What mass of ethylene glycol (C2H6O2) is required to lower the freezing point of 50.0 g of water by 10.0oC?

Homework Answers

Answer #1

Freezing Point Depression

(change in Freezing point) ΔTF = KF · b

ΔTF, = TF (pure solvent) - TF (solution) = 10.0oC

KF, the cryoscopic constant, KF = 1.853 oC /m

b, is the molality (mol solute per kg of solvent)

b = mole solute / 0.05 kg

then;

ΔTF = KF · b

10.0oC = 1.853 oC /m. mole solute / 0.05 kg

Number of moles of solute = (10.0*0.05 )/1.853

= 0.27 mol

Now convert it into grams by molar mass of solute means ethylene glycol (C2H6O2)

= 0.27 mol * 62.07 g/mol

=16.75 g ethylene glycol (C2H6O2)

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