Question

What is the freezing point of a solution prepared from 45.0 g ethylene glycol (C2H6O2) and 85.0 g H2O? Kf of water is 1.86°C/m.

Answer 1

Answer:

$T_{sol}=-15.9\°C$

Explanation:

Hello,

In this case, we can analyze the colligative property of solutions - freezing point depression - for the formed solution when ethylene glycol mixes with water. Thus, since water freezes at 0 °C, we can compute the freezing point of the solution as shown below:

$T_{sol}=T_{water}-i*m*Kf$

Whereas the van't Hoff factor for this solute is 1 as it is nonionizing and the molality is:

$m=\frac{mol_{solute}}{kg\ of\ water}=\frac{45.0g*\frac{1mol}{62g} }{85.0g*\frac{1kg}{1000g} } =8.54m$

Thus, we obtain:

$T_{sol}=0\°C+(-8.54m*1.86\frac{\°C}{m} )\\\\T_{sol}=-15.9\°C$

Best regards.