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

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.