Question

An ethylene glycol solution contains 29.8 g of ethylene glycol (C2H6O2) in 92.6 mL of water. (Assume a density of 1.00 g/mL for water.)

Part A

Determine the freezing point of the solution.

Express you answer in degrees Celsius.

Part A

Determine the freezing point of the solution.

Express you answer in degrees Celsius.

Part B

Determine the boiling point of the solution.

Express you answer in degrees Celsius.

Answer 1

Part A:-

We know that ΔT _f = K_f x m
Where

ΔT _f = depression in freezing point

        = freezing point of pure solvent – freezing point of solution

        = 0 - T_f   ^oC

       = - T_f   ^oC

K _f = depression in freezing constant of water = 1.86 ^oC/m

m = molality of the solution

    = ( mass / Molar mass ) / weight of the solvent in Kg

    = (29.8 / 62) / 0.0926 kg

    = 5.19 m

Plug the values we get    - T_f = 1.86x5.19

= 9.65 ^oC

                                     T_f = -9.65 ^oC

So the freezing point of the solution is -9.65 ^oC

Molar mass of C2H6O2 = (2x12) + (6x1) + (2x16) = 62 g/mol

Mass of water = density x volume

                     = 1.00(g/mL) x 92.6 mL

                     = 92.6 g

Part B:-

We know that ΔT _b = K_bx m
Where

ΔT _b= elevation in boiling point

        = boiling point of solution - boilinging point of pure solvent

        = T_b -100 ^oC

K _f = elevation in boiling point constant of water = 0.152 ^oC/m

m = molality of the solution

    = ( mass / Molar mass ) / weight of the solvent in Kg

    = (29.8 / 62) / 0.0926 kg

    = 5.19 m

Plug the values we get

T_b -100 = 0.512x5.19

           = 2.66

T_b =100+2.66

    = 100.26 oC