An aqueous solution of potassium hydroxide (KOH) is fed to an evaporative crystallizer at a rate of 875 kg/h. The crystallizer operates at 10°C and produces crystals of KOH•2H₂O. Water evaporated from the crystallizer flows to a condenser, and the resulting condensate is collected in a tank. During a 30-minute period, 73.8 kg of water is collected. Five-gram samples of the feed to the crystallizer and the liquid removed with the crystals are taken for analysis and subsequently titrated with 0.85 M H₂SO₄. It is found that 22.4 mL of the H₂SO₄ solution is required for the feed and 26.6 mL is required for the product liquid. 

(a) What fraction of the KOH in the feed is crystallized? 

(b) Later you learn that a solution in equilibrium with KOH•2H₂O crystals at 10°C has a concentration of 103 kg KOH/100 kg H₂O. How would this information cause you to reconsider the procedure by which a sample of the mother liquor was obtained? (Consider removing a slurry sample—i.e., one containing both solution and KOH•2H₂O crystals—that is maintained at 10°C, but that initially had a solute concentration of 121 kg KOH/100 kg H₂O. What would that concentration be after the sample is stored for several hours?)