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.
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 answer
1 answer