Answer:
[tex]T_{sol}=-15.9\°C[/tex]
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:
[tex]T_{sol}=T_{water}-i*m*Kf[/tex]
Whereas the van't Hoff factor for this solute is 1 as it is nonionizing and the molality is:
[tex]m=\frac{mol_{solute}}{kg\ of\ water}=\frac{45.0g*\frac{1mol}{62g} }{85.0g*\frac{1kg}{1000g} } =8.54m[/tex]
Thus, we obtain:
[tex]T_{sol}=0\°C+(-8.54m*1.86\frac{\°C}{m} )\\\\T_{sol}=-15.9\°C[/tex]
Best regards.
The freezing point of a solution prepared from 45.0 g ethylene glycol and 85.0 g of water is -15.9 °C.
Freezing point depression is a drop in the temperature at which a substance freezes, caused when a smaller amount of another, non-volatile substance is added.
We will use the definition of molality.
b = mass of solute / molar mass of solute × kg of solvent
b = 45.0 g / 62.07 g/mol × 0.0850 kg = 8.53 m
We will use the following expression, where Kf is the cryoscopic constant of water.
ΔT = Kf × b = 1.86 °C/m × 8.53 m = 15.9 °C
The freezing point of pure water is 0°C.
T = 0°C - 15.9 °C = -15.9 °C
The freezing point of a solution prepared from 45.0 g ethylene glycol and 85.0 g of water is -15.9 °C.
Learn more about freezing point depression here: https://brainly.com/question/14115775
