Question

You are required to prepare 100 gm of crystals of pure potassium nitrate from its saturated solution at 75°C. The room temperature is 20°C and the solubility of potassium nitrate at 75°C and 20°C is 155 gm and 31 gm per 100 g of water respectively. Find the initial weight of potassium nitrate required to achieve it, assuming water available is 100 gm.

Answer 1

• The solubility of potassium nitrate at 75°C is 155 gm per 100 g and solubility at 20°C is 31 gm per 100 g.
• So for 100 g of a saturated solution that contains dissolved KNO₃ at 75 °C.
• $100gofsolution\times \frac{155gKN{O}_3}{(100+100)gsolution}=60.78gKN{O}_3$
• M_{water =}100-60.78 =39.22 g.
• KNO₃ can be dissolved in 39.22 g of water at 25°C to make a saturated solution,
• Maximum amount of dissolved KNO₃ $39.22gofwater\times \frac{31gKN{O}_3}{100gKN{O}_3}=12.15$.
• When the initial solution is cooled from 75°C to 25°C the amount of water that it contained will only hold 12.15 g of dissolved KNO₃.
• The rest will crystallize out of the solution
• 60.78 g - 12.15 g = 48.68 g
• Hence the initial weight of potassium nitrate is 48.86 g.