Automotive antifreeze consists of ethylene glycol, CH2(OH)CH2(OH), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0% mass solution of ethylene glycol in water. (10 points) (For H2O, kb = 0.51 °C/m, kf = 1.86 °C/m)

Freezing point of solution =  - 10.3°C

Boiling point of solution  = 102.8°C

Explanation:

This is about colligative properties:

Freezing point depression → ΔT =  Kf . m

ΔT =  Freezing point of pure solvent - Freezing point of solution

Boiling point elevation → ΔT = Kb . m

ΔT =  Boiling point of solution - Boiling point of pure solvent

Let's determine m which means molality (moles of solute in 1 kg of solvent)

Solute: Ethylene glycol → Mass = 25 g

Moles = Mass / Molar mass → 25 g / 60g/mol = 0.416 moles

Solvent: Water

25 % means, 25 g of solute in 100 g of solution

Therefore, the mass of water is 75 g (100 -25) . (Solution = Solute + Solvent)

We convert the mass from g to kg → 75 g . 1kg /1000g = 0.075 kg

Molality (mol/kg) = 0.416 m / 0.075kg = 5.55 m

We replace data in the formulas:

Freezing point depression:0° - Freezing point of solution= 1.86°C/m. 5.55 m

Freezing point of solution = - (1.86°C/m . 5.55m) = - 10.3°C

Boiling point elevation:Boiling point of solution - 100°C = 0.51°C /m . 5.55 m

Boiling point of solution = 0.51°C /m . 5.55 m + 100°C = 102.8°C

hurricanes form when warm, moist air from the ocean surface quickly rises into the atmosphere and forms storm clouds and rain. when the clouds are formed, heat is released which warms the cool air in the upper regions of the atmosphere so that the cycle can continue. which two phases of the water cycle are repeated in this process?