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) + CO₃^2-(aq)

Every mol Li₂CO₃ that dissolves produces 3 mol of particles.

Calculate the change in freezing point;

T_fp = ik_fpm

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

T_fp = 0.544 degrees C

T_fp = -0.544 degrees C

Calculate the T_bp and check your answer.

T_fp = ik_bpm

T_bp = 3 mol particles ( 0.512 degrees C m^-1) 0.0974 molal

T_bp = 0.149 degrees C

T_bp = 100.149 degrees C

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