Question

Several amount od unknown salt was added to a solvent. Initial amount of the solvent is 30ml. When the mixture weighed 0.2202 lb, undissolved salts were observed. Upon separation of the salts, it was found to be around 5 g. What is the solubility of the unknown salt? Before separating the undissolved solute from the mixture, was the solution supersaturated and why? (Density of the solvent = 1.02 g/mL)

Answer 1

0.2202 lb = 99.88 g

m(solvent) $= 30 \times 1.02 = 30.6\; g$

∆m = 99.88 – 30.6 – 5 = 64.28 g (mass of dissolved salt)

Solubility = 64.28 g/ 30 mL = 2.142 g/mL = 2142 g/L

Yes, solution was supersaturated, because concentration of a solute exceeds the concentration specified by the value equilibrium solubility.