Final answer:
To dissolve 2.0 g of impure salicylic acid for recrystallization, approximately 0.19 mL of boiling water is needed. After cooling, the maximum amount of solid salicylic acid isolated after vacuum filtration is approximately 1430 mL.
Step-by-step explanation:
To calculate the minimum amount of boiling water necessary to dissolve 2.0 g of impure salicylic acid for recrystallization, you need to convert the mass of salicylic acid into moles using its molar mass. Then, you can use the solubility data to determine the volume of boiling water needed. The molar mass of salicylic acid is 138.1 g/mol, so 2.0 g is equivalent to 2.0/138.1 = 0.0145 mol. Using the solubility at 100°C, 77.8 g/L, you can calculate the volume of boiling water needed as 0.0145 mol / (77.8 g/L) = 0.00019 L or 0.19 mL.
When the solution is cooled to room temperature and then to 10°C with an ice/water bath, the solubility of salicylic acid decreases. According to the solubility data, at 10°C the solubility is 1.4 g/L. To determine the maximum amount of solid salicylic acid isolated after vacuum filtration, you need to calculate the volume of water that can hold 2.0 g of salicylic acid at 10°C. Using the solubility at 10°C of 1.4 g/L, you can calculate the maximum volume of water as 2.0 g / (1.4 g/L) = 1.43 L or 1430 mL.