Question

What is the boiling point of a solution containing 203 g of ethylene glycol (C2H6O2) and 1035 g of water? (Kb for water is 0.52 oC/m).

Answer 1

101,37°C

Step-by-step explanation:

Boiling point elevation is one of the colligative properties of matter. The formula is:

ΔT = kb×m (1)

Where:

ΔT is change in boiling point: (X-100°C) -X is the boiling point of the solution-

kb is ebulloscopic constant (0,52°C/m)

And m is molality of solution (mol of ethylene glycol / kg of solution). Moles of ethylene glycol (MW: 62,07g/mol):

203g × (1mol /62,07g) = 3,27moles of ethlyene glycol

Molality is: 3,27moles of ethlyene glycol / (1,035kg + 0,203kg) = 2,64m

Replacing these values in (1):

X - 100°C = 0,52°C/m×2,64m

X - 100°C = 1,37°C

X = 101,37°C

I hope it helps!