5. The following table gives the solubility of an unknown organic compound in wa

Question

5. The following table gives the solubility of an unknown organic compound in water at several different temperatures: Solubility of unknown compound in water/)Temperature(C) 0.752 2.34 45.7 0 25 100 a. What is the minimum amount of boiling water (in ml:'s) required to fully dissolve 1.00 g of the unknown compound? [1 point] b. A student recrystallizes 1.00 g of the unknown compound from 50 mL of water. After allowing the mixture to cool to room temperature (25 C), the student collects the recrystallized compound using vacuum filtration. What is the theoretical maximum percent recovery of the unknown compound for this recrystallization step? (3 points) c. A student recrystallizes 1.00 g of the unknown compound from 50 mL of water. After allowing the mixture to cool in an ice bath (0 C), the student collects the recrystallized compound using vacuum filtration. What is the theoretical maximum percent recovery of the unknown for this recrystallization step? 13 points) d. The solubility of the same unknown organic compound in ethanol (CHsCH 0H) is 4.52 B/L at 25 *C. Would ethanol be a better recrystallization solvent than water for this particular unknown compound? Briefly explain your answer. [2 points

Explanation / Answer

Solution :-

Part a)

At 100 degree C the solubility is 45.7 g/L

That means 45.7 g solid dissolve in 1000 mL water

Lets calculate the volume of boiling water needed to dissolve 1.00 g unknown solid

1.0g * 1000 ml / 45.7 g = 21.9 mL

Therefore it will need 21.9 mL of boiling water.

Part b) At 25 degree C solubility of the compound is 2.34 g/L

Student recrystalizes 1.00g compound in 50 ml water

Lets calculate the amount of unknown soluble in 50 ml (0.050 L) water at 25 C

(2.34 g / 1 L)*0.050 L = 0.117 g

Therefore out of the 1.00 g compound 0.117 g is soluble at 25 C therefore the amount of recrystallized solid that can be recovered is

1.0g – 0.117 g= 0.883 g

% recovery = [mass of solid recovered / mass of solid recrystalized]*100%

                    = [0.883 g /1.0 g] *100 %

                    = 88.3 %

Part c)

At 0 degree C solubility of the compound is 0.752 g/L

Student recrystalizes 1.00g compound in 50 ml water

Lets calculate the amount of unknown soluble in 50 ml (0.050 L) water at 25 C

(0.752 g / 1 L)*0.050 L = 0.0.0376 g

Therefore out of the 1.00 g compound 0.0376 g is soluble at 0 C therefore the amount of recrystallized solid that can be recovered is

1.0g – 0.0376 g= 0.9624 g

% recovery = [mass of solid recovered / mass of solid recrystalized]*100%

                    = [0.9624 g /1.0 g] *100 %

                    = 96.24 %

Part d) solubility of the unknown compound is more in ethanol than in water at 25 degree C therefore at room temperature more unknown compound will remain dissolved in the solvent and will not crystalizes out therefore it is not good choice of solvent.

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