Question

An ethylene glycol solution contains 30.0g of ethylene glycol (C2H6O2) in 95.6mL of water. (Assume a density of 1.00 g/mL for water.)

1.Determine the freezing point of the solution.

Express you answer in degrees Celsius

2.Determine the boiling point of the solution.

Express you answer in degrees Celsius.

Answer 1

first calculate the molality of the solution

molality =(W/MW ) (1000 / mass of solvent in g)

molality = (30.0 / 62.07) (1000 / 95.6)

molality = 5.05 m

now

$\Delta$ Tf = Kf x m

$\Delta$ Tf = 1.86 x 5.05

$\Delta$ Tf = 9.40

freezing point = 0 - 9.40 = -9.40

freezing point = - 9.40 ⁰C

2) $\Delta$ Tb = Kb x m

$\Delta$ Tb = 0.512 x 5.05

$\Delta$ Tb = 2.58

boiling point = 100 + 2.58

boiling point = 102.58 ⁰C

freezing point will decrease so we substract.

boiling point will increase so we are adding.