Question

It takes 6.86 kg of ethylene glycol (antifreeze) to decrease the freezing point of 6.50 kg of water to -25.0ºF (-31.7ºC). How much lithium carbonate (Li2CO3) would it take to decrease the freezing point of 7.10 kg of water to -25.0ºF? (Assuming all the salt will dissolve in that amount of water.)

1)

5.68 kg

2)

8.95 kg

3)

4.48 kg

4)

2.98 kg

5)

2.35 kg

Answer 1

Ans. Given, mass of ethylene glycol = 6.86 kg = 6860.0 g

Moles of ethylene glycol = Mass / Molar mass

                                                = 6860.0 g / (62.07 g/mol)

                                                = 110.520 mol

Molarity of solution = Moles of ethylene glycol / Mass of water in kg

                                    = 110.520 mol / 6.50 kg

                                    = 17.00 mol/ kg                                          ; [1 mol/ kg = 1 m]

                                    = 17.00 m

Now, freezing point depression, dT_f is given by-

            dT_f = i K_f m               - equation 1

            where, i = Van’t Hoff factor. [i = 1 for non-electrolyte solute]. Since lauric acid is a non-polar solvent, it’s assumed that the unknown is also non-polar. So, i = 1.

                        K_f = molal freezing point depression constant of solvent = 3.9⁰C / m

                        m = molality of the solution

                        dT_f = Freezing point of pure solvent – Freezing point of solution

Putting the values in equation 1-

            0.0⁰C – (- 31.7⁰C) = 1 x Kf x 17.00 m

            Or, 31.7⁰C = Kf x 17.00 m

            Or, Kf = 31.7⁰C / 17.00 m = 1.864⁰C/m

            Hence, Kf = 1.864⁰C/ m

[The experimental value of Kf is same as theoretical one].

# Calculating the amount of Li₂CO₃

Li₂CO₃ dissociates into 2 Li-ions and one carbonate ion when dissolved in water.

Thus, i = 3 for Li₂CO₃.

Let the molarity of solution be A.

Putting the values in equation 1-

            0.0⁰C – (- 31.7⁰C) = 3 x Kf x A                    ; [Kf = 1.864⁰C/ m –as calculated above]

            Or, 31.7⁰C = 3 x (1.864⁰C/ m) x A

            Or, A = 31.7⁰C / (5.592⁰C / m) = 5.669 m

            Hence, A = 5.669 m

Therefore, molarity of specified solution = 5.669 m

Now,

Moles of Li₂CO₃ in 7.10 kg water = Molarity x Mass of water in kg

                                                = 5.669 m x 7.10 kg

                                                = (5.669 mol/ kg) x 7.10 kg

                                                = 40.249 mol

Mass of Li2CO3 required = Moles x Molar mass

                                                = 40.249 mol x (73.8912 g/mol)

                                                = 2974.04 g

                                                = 2.97 kg

Thus, correct option is- (4). 2.98 kg [nearest value].