Question

K_f, water = 1.86 K Kg mol^-1, K_f camphor = 40.0 K Kg mol^-1 Estimate the freezing point of aqueous solutions containing 10.26 g of sucrose (nonelectrolyte) in 100 g of water 1.75 g of sodium chloride (strong electrolyte) in 100 g of water 6.25 g of barium chloride (strong electrolyte) in 100 g of water

Answer 1

Question 2 a

Formula = Δt = i K_f m

i = van 't Hoff factor

Kf = molal freezing point depression constant

Δt = The temperature change

m = molality of the solution

Determine the molality of sucrose = 10.26 / 342.29 = 0.0299 /0.1 kg water = 0.299 m

Determine the freezing point depression:

Δt = i K_f m

Δt = 1 x 1.86 x 0.299 = 0.5575 K

Determine the freezing point:

0 - 0.5575 = - 0.5575 °C

Question 2

Δt = i K_f m

In this case i = 2 because in water it will dissociate to sodium ions and chloride ions.

Determine the molality of NaCl = 1.75 / 58.45 = 0.0299 /0.1 kg water = 0.299 m

Determine the freezing point depression:

Δt = 2 x 1.86 x 0.299 = 1.113 K

Determine the freezing point:

0 - 1.113 = - 1.113   °C

Question 3

In this case i = 3 because in water it will dissociate to one barium ion and 2 chloride ions.

Determine the molality of BaCl2 = 6.25 / 208.23 = 0.03 /0.1 kg water = 0.3 m

Determine the freezing point depression:

Δt = 3 x 1.86 x 0.3 = 1.674 K

Determine the freezing point:

0 - 1.674 = - 1.674 °C