FOR 2-3 Solubility data for compound A 223 g/500 mL of water at 100oC 75.1 g/500
ID: 1022033 • Letter: F
Question
FOR 2-3
Solubility data for compound A 223 g/500 mL of water at 100oC
75.1 g/500 mL of water at 5oC
2)What is the minimum volume of water required to recrystallize a 431 g sample of A?Please carry maximum number of significant figures through to final answer then round to a whole number. Do not include units.
3)Assuming a student used 2125 mL of water in question 2 what is the maximum percent recovery of A given that the solution was cooled to 5oC? Please carry maximum number of significant figures through to final answer then round to a whole number.
4)Rank compounds B, C and D from most soluble in water to least soluble in water (most soluble in water first).
[1.5 pts] What quantity of 6.75 M aqueous sodium hydroxide? Ple uld be required to neutralize 265 mL of 8.40 M significant figures through to final answer then 1. ase Press F11 to exit full screen nd to a whole numb 1.5 pts] What is the minimum volume of water required to recrystallize a 431 g sample of A? Please carry maximum number of significant figures through to final answer then round to a whole number. Do not include units. 2. Solubility data for 223 g/500 mL of water at 100°C 75.1 g/500 mL of water at 5°C 3. 1.5 pts] Assuming a student used 2125 mL of water in question 2 what is the maximum percent recovery of A given that the solution was cooled to 5°C? Please carry_maximum number of significant figures through to final answer then round to a whole number. Do not include units lease carry maximum number of significant [1.5 pts] Rank compounds B, C and D from most soluble in water to least soluble in water (most soluble in water first). 4. OD> B>C OHExplanation / Answer
1) Write down the molecular equation for neutralization:
2 NaOH (aq) + H2SO4 (aq)--------> Na2SO4 (aq) + 2 H2O (l)
Moles of NaOH in the neutralization reaction = (265mL)*(1 L/1000 mL)*(8.40 mole/L) = 2.226 moles.
As per the molecular equation above, the molar ratio of NaOH:H2SO4 = 2:1; therefore, moles of H2SO4 required for the neutralization reaction = (2.226 mole NaOH)*(1 mole H2SO4/2 mole NaOH) = 1.113 mole H2SO4.
The molar concentration of H2SO4 is 6.75 M; let the volume required be x L.
Therefore,
(x L)*(6.75 mole/L) = 1.113 mole
===> x = 0.164889
The volume of H2SO4 required (in mL) = (0.164889 L)*(1000 mL/1 L) = 164.889 mL 164.9 mL
Ans: 164.9 or if you round off further, then 165.0 (possibly)
2) 500 mL water contains 223 gm compound A at 100°C.
Hence, 100 mL water will contain (223 gm)*(100 mL water/500 mL water) = 44.6 gm compound A.
Thus, the solubility of the compound is 44.6 gm/100 mL water at 100°C.
Let the volume of water required be x mL.
Therefore,
44.6 gm/100 mL = 431 gm/x mL
===> x = (431*100)/44.6 mL = 966.367 mL 966.4 mL
Ans: 966.4
3) We have 75.1 gm compound A in 500 mL water at 50°C.
Therefore, 100 mL water will contain (75.1 gm)*(100 mL water/500 mL water) = 15.02 gm compound A.
Therefore, solubility at 50°C = 15.02 gm/100 mL water
2125 mL water was used; let the amount remaining in water be y gm at 50°C.
Hence, 15.02 gm/100 mL = y/2125 mL
===> y = 15.02*2125/100 gm = 319.175 gm
Therefore, 319.175 gm compound A remains in solution.
Amount recovered = (431 – 319.175) gm = 111.825 gm
Percent recovery = (111.825/431)*100 = 25.9454 25.95
Ans: 25.95
4) The correct option is C>D>B.
C is the sodium salt of an acid. Since it contains an ion pair (carboxylate anion and sodium cation), it will be the most soluble in water.
D contains a hydroxylic group which is polar in nature. However, there are double bonds and a bicyclic ring which are non-polar, hence it is less polar than C.
B is a bicyclic arene with no polar groups; hence it will be least polar and hence least soluble in water.
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