Question

To 5.00 L of water at 5.0 ° C is added 724.43 g of calcium chlorate and 1045.5 g of potassium sulfate. After the precipitation of an insoluble salt, filtered to remove the precipitated solid.
a) What mass of solid has been removed?
b) What is the freezing temperature of the filtered aqueous solution?

Expected answers :

a) mass of solid (calcium sulphate) = 476 g

b) -8.00 ° C.

Answer 1

a) Ca(ClO3)2 + K2SO4 ------> CaSO4(s) + 2KClO3(aq)

given mole of Ca(ClO3)2 = 724.43g/206.98g/mol = 3.5

given mole of K2SO4 = 1045.5g/174.29g/mol = 5.9986

stoichiometrically, 3.5 mole of Ca(ClO3)2 require 3.5mole of K2SO4 but available K2SO4 is 5.9986

So, Ca(ClO3)2 ia limiting reagent

3.5 mole of Ca(ClO3)2 will form 3.5mole of CaSO4

mass of CaSO4 formed = 3.5mol × 136.14g/mol = 476g

b) mole of K2SO4 dissolved = 5.9986 - 3.5 =2.4986mol

molality = 2.4986mol/5kg = 0.4997m

mole of KClO3 dissolved = 3.5 ×2 = 7 mol

molality = 7mol/5kg = 1.4m

Freezing point depression, ∆T_F = K_F × b × i

K_F = Freezing point depression constant , 1.86℃/m

b= molality of solute

i = vant hoff factor, for K2SO4 = 3 , for KClO3 = 2

∆T_F for KClO3

∆T_F = 1.86℃/m × 1.4m × 2

= 5.208℃

∆T_F for K2SO4

∆T_F = 1.86℃/m × 0.4997m × 3

= 2.788℃

Total ∆T_F = 5.208℃ + 2.788℃ = 8.00℃

∆T = Freezing point of water - Freezing point of solution

Freezing point of water - Freezing point of solution= 8.00℃

0℃ - Freezing point of solution = 8.00℃

Freezing point of solution = -8.00℃