Question

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

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)