Idont understand it 1.what is the boiling point of a solution of 675 grams of ethylene glycol (c2h6o2) in 2.50 liters of water?

102 °C

Explanation:

The boiling point elevation is computed as follows:

ΔT = k*m*i

where ΔT is the difference between the solution's boiling point and water's boiling point, k is a constant (0.51 °C*kg/mol for water), m is the molality of the solution, and i is the van't Hoff constant (equal to 1 for ethylene glycol  because is a covalent molecule and it doesn't dissociate).

Molar mass of ethylene glycol: 62.07 g/mol

Moles of ethylene glycol: mass / molar mass = 675/62.07 = 10.87 mol

Density of water: 1 kg/L

Mass of water: density * volume = 1*2.5 = 2.5 kg

Molality of the solution: moles of solute / kg of solvent = 10.87/2.5 = 4.348 mol/kg

Then:

ΔT = 0.51*4.348*1 = 2.217 °C

Boiling point of water: 100 °C

Then, the boiling point of the solution is 100 + 2.217 ≈ 102 °C