Question

The solubility of benzoic acid in water at 95C is 68g/L. At room temperature (25C), the solubility of benzoic acid in water is 3.4 g/L

A. What is the minimum amount of water in which one gram of benoic acid can be dissolved at 95C?
B. based upon your answer to part a, when this 95C solution is cooled to 25C, how much benzoic acid will remain in solution and how much will precipitate/crystallize out of the solution?

Answer 1

(A) Given the solubility of benzoic acid in water at 95^oC is 68g/L

So 1L of water requires 68 g of benzoic acid

     M L of water requires 1 g of benzoic acid

     M = ( 1x1) / 68

        = 0.0147 L

        = 14.7 mL                 since 1 L = 1000 mL

Therefore the amount of water required is 14.7 mL

(B) Given the solubility of benzoic acid in water at 25^oC is 3.4g/L

So 1L of water requires 3.4 g of benzoic acid

     0.0147 L of water requires N g of benzoic acid

   N = (0.0147 x 3.4) / 1

      = 0.05 g

So 0.05 g of benzoic acid will remain in solution.

Therefore 1.00-0.05 = 0.95 g of benzoic acid will precipitate