plz help! don't understand problem at all.
The solubility profile of tartaric acid in water is given in the following table.
Temperature (C) Solubility (g/100 mL of water)
0 115
10 126
20 139
30 156
40 176
50 195
60 217
70 244
80 273
90 307
1. Suppose 1.0 g of tartaric acid and 1.0 mL of water were mixed and heated to 90°C. Would all of the tartaric acid dissolve? Explain.
2. The solution prepared in (a) is cooled. At what approximate temperature will crystals of tartaric acid first appear?
3. Suppose the cooling described in (b) were continued to 0°C. How many grams of tartaric acid would come out of solution? Show calculation to justify your answer.
1. (307 g/100 mL) x 1 mL = 3.07g so 1 g will completely dissolve.
2. approximately zero but a little less than zero.
3. (115 g/100) x 1 mL = 1.15 so none will come out of solution since it can hold 1.15 and you have only 1.0 g.
Sure, let's go through each question step by step:
1. To determine if all of the tartaric acid would dissolve at 90°C, we need to compare the solubility of tartaric acid at that temperature (307 g/100 mL of water) to the amount of tartaric acid in the mixture (1.0 g). Since the solubility is given in grams per 100 mL, we need to convert the volume of water to milliliters. 1.0 mL is equivalent to 1.0 mL/100 mL = 0.01 (10^-2) of the total volume.
Therefore, the amount of tartaric acid in the mixture relative to the solubility can be calculated as follows:
(1.0 g / 100 mL) * (0.01 * 100 mL) = 0.01 g
Since 0.01 g is less than the solubility at 90°C (307 g/100 mL), we can conclude that all of the tartaric acid would dissolve at 90°C.
2. To determine the approximate temperature at which crystals of tartaric acid will first appear when the solution from part (a) is cooled, we need to find the temperature at which the solubility of tartaric acid is equal to or lower than the amount of dissolved tartaric acid. In this case, the solubility should be equal to the dissolved amount since any additional cooling would cause crystallization.
From the table, we can see that the solubility at 0°C is 115 g/100 mL. Using the same calculation as before, we can find the amount of tartaric acid relative to the solubility:
(1.0 g / 100 mL) * (0.01 * 100 mL) = 0.01 g
Therefore, when the solubility is equal to 0.01 g, crystals of tartaric acid will start to appear. Based on the table, this occurs at around 60°C (solubility of 217 g/100 mL).
3. To determine how many grams of tartaric acid would come out of solution when the solution from part (b) is cooled to 0°C, we need to find the difference in solubility between 60°C and 0°C.
From the table, we can see that the solubility at 60°C is 217 g/100 mL, and the solubility at 0°C is 115 g/100 mL. The difference is 217 g/100 mL - 115 g/100 mL = 102 g/100 mL.
Since the original volume of water was 1 mL, we can calculate the amount of tartaric acid that would come out of solution:
(102 g / 100 mL) * (1 mL / 100 mL) = 1.02 g
Therefore, when the solution is cooled to 0°C, approximately 1.02 grams of tartaric acid would come out of solution.