The solubility of compound A in ethanol is 0.5 g per 100 mL at 0oC and 5.0 g per 100 mL at 78oC.
(a)
To make a hot, saturated solution of 15.0 g of A at 78oC, how much ethanol is required?
(b)
When the solution prepared in is cooled to 0oC, how much of compound A will be crystallized? How much of A would remain in solution and would not be recoverable?
a.
100 mL x 15 g/5.0 g = 300 mL ethanol required to prepare a saturated solution.
b. After cooling this to 0C,
0.5 g x (300 mL/100 mL) = 1.5 g that will be lost due to solubility.
Amount recovered will be 15.0-1.5 = ? recrystallized.
To answer both parts (a) and (b) of the question, we will need to use the given solubility data for compound A in ethanol at 0°C and 78°C.
(a) To make a hot saturated solution of 15.0 g of compound A at 78°C, we need to determine how much ethanol is required.
From the given data, we know that at 78°C, the solubility of compound A in ethanol is 5.0 g per 100 mL. Therefore, for every 5.0 g of compound A, we need 100 mL of ethanol.
To find the amount of ethanol needed for 15.0 g, we can use the following calculation:
(15.0 g) x (100 mL / 5.0 g) = 300 mL
So, to make a hot saturated solution of 15.0 g of compound A at 78°C, you will need 300 mL of ethanol.
(b) When the solution prepared in part (a) is cooled to 0°C, we need to determine how much of compound A will crystallize and how much will remain in solution.
At 0°C, the solubility of compound A in ethanol is 0.5 g per 100 mL.
First, let's calculate the maximum amount of compound A that can be dissolved in 300 mL of ethanol at 0°C:
(0.5 g / 100 mL) x 300 mL = 1.5 g
So, at 0°C, the maximum amount of compound A that can be dissolved in 300 mL of ethanol is 1.5 g.
Since 15.0 g of compound A were originally dissolved in the hot solution, and only 1.5 g can be dissolved at 0°C, the excess compound A will crystallize out of the solution.
The amount of compound A that will crystallize out can be calculated as the difference between the original amount (15.0 g) and the amount remaining in solution (1.5 g):
15.0 g - 1.5 g = 13.5 g
So, when the solution prepared in part (a) is cooled to 0°C, 13.5 g of compound A will crystallize out of the solution.
The remaining 1.5 g of compound A will stay dissolved in the 300 mL of ethanol, and it cannot be recovered unless further separation techniques are used.