1. The table below shows the solubility of a certain organiccompound in water at
ID: 689990 • Letter: 1
Question
1. The table below shows the solubility of a certain organiccompound in water at five different temperatures.0 20 40 60 80 Temperature (oC) 1.5 g 3.0 g 6.5 g 11.0g 17.0g Solubility of compound (in 100 mL of water)
b) If a student attempts to recrystallize a 0.5 g sample ofthis compound by heating it to 80 C with 5.0 mL of water, would allof the sample dissolve? Briefly justify your answer.
c) Assuming that the answer to part b is ‘Yes’, atwhat temperature will the crystals begin to appear when thestudent’s solution begins to cool?
d) If the student cooled the solution to 0 C and filtered offthe crystals, what is the maximum possible percentage recovery?What mass of the sample will remain in the filtrate? 1. The table below shows the solubility of a certain organiccompound in water at five different temperatures.
0 20 40 60 80 Temperature (oC) 1.5 g 3.0 g 6.5 g 11.0g 17.0g Solubility of compound (in 100 mL of water)
b) If a student attempts to recrystallize a 0.5 g sample ofthis compound by heating it to 80 C with 5.0 mL of water, would allof the sample dissolve? Briefly justify your answer.
c) Assuming that the answer to part b is ‘Yes’, atwhat temperature will the crystals begin to appear when thestudent’s solution begins to cool?
d) If the student cooled the solution to 0 C and filtered offthe crystals, what is the maximum possible percentage recovery?What mass of the sample will remain in the filtrate?
Explanation / Answer
b) Determine what concentration is possible at 80 C: At 80 C, 17.0 g/100 mL water = 0.17 g/mL. 0.5 g/5.0 mL = 0.10 g/mL Since the concentration in part b is smaller than the possibleconcentration at 80 C, all of the sample will dissolve. c) The crystals will begin to appear when the temperature drops toa point where the possible concentration at that temperature isless than 0.10 g/mL. At 60 C the possible concentration is0.11 g/mL which is above 0.10 g/mL; so at 60 C no crystals willform. At 40 C the possible concentration is 0.065 g/mL whichis less than 0.10 g/mL; so the crystals will have already appearedby this temperature. The temperature is somewhere between 40and 60 C. Since the relationship between temperature andpossible concentration is not linear, you cannot know the exacttemperature where crystals will appear from this data. Justguessing from the data, the actual temperature is probably around55-59 C. d) At 0 C the concentration will be 0.015 g/mL. You are using5.0 mL; so you will lose 0.015 g/mL * 5.0 mL = 0.075 g left in thefiltrate. The amount you will recover is 0.5 g - 0.075 g =4.925 g. The percentage recovered will be: (4.925 g/5.0 g) * 100% = 98.5% Hope this helps. :-)
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