Calculate the freezing point and the boiling point of a saturated solution of Li2CO3. The solubility of lithium carbonate is 0.72 g per 100 g of water at 100 degrees C.

Answer:

We need to begin by calculating the molality of the saturated Li2CO3 solution.

We know the molality of the Li2CO3 solution, but we need to consider that Li2CO3 is a strong electrolyte. Determine the number of particles Li2CO3 forms when dissolved in water.

Li2CO3(s) -H2O--> 2Li+(aq) + CO32-(aq)

Every mol Li2CO3 that dissolves produces 3 mol of particles.

Calculate the change in freezing point;

Tfp = ikfpm

Tfp = 3 mol particles ( 1.86 degrees C m-1) 0.0974 molal

Tfp = 0.544 degrees C

Tfp = -0.544 degrees C

Calculate the Tbp and check your answe here.