Calculate the freezing point and boiling point of an antifreeze solution that is 49.6% by mass ethylene glycol (hoch₂ch₂oh) in water. ethylene glycol is a nonelectrolyte.

- Freezing point of solution:

-28.03°C

- Boiling point of solution: 108.06°C

Explanation:

49.6% by mass means 49.6 g of solute in 100g of solution. To work with the colligative properties with must use molality (mol of solute / 1 kg of solvent). We should know the moles of ethylene glycol (mass / molar mass).

49.6 g / 62.07 g/mol = 0.799 moles

The mass of solvent is:

100 g - 49.6 g = 51.4 g

Now we calculate molality:

0.799 m / 0.0514 kg = 15.5 m.

We can work with the colligatives now.

ΔT = Kf . m . i

Where ΔT = T° frezzing pure solvent - T° freezing solution

Kf is cryoscopic constant

m is molality

i = Vant Hoff factor (number of ions dissolved, as ethylene glycol is non eletrolytic, i = 1).

So the t° freezing of solution is:

0°C - T° f sol. = 1.86 °C/m . 15.5m . 1

T° freezing sol = - 28.83°C

Let's go to boiling point. Formula is:

ΔΤ = Kb . m .i

ΔΤ = T° boiling of solution - T° boiling pure solvent

Kb = Boiling consant for water

m = molality

T° boiling solution - 100°C = 0.52°C/m . 15.5 m . 1

T° boiling solution = 0.52°C/m . 15.5m + 100°C =108.06°C

ok whats the question if not i cannot you i need question and need a thing to answer