posted by Taylor on .
I don't get two questions out of my homework and I know the answers to them but just not how the answers were gotten. PLEASE HELP!!
1) A handbook gives the aqueous solubility of carbon monoxide at 0 C and 1 atm CO pressure to be 0.0354 mg CO per mL of solution. What would be the molarity of CO in water if the partial pressure of CO is 0.00036 atm? ANS. is 4.5*10^-7
2) The solubilities of ammonium bromide in water is as follows:
T (C) Sol. g NH4Br/100g H20
Which of the following fractional crystallization schemes would produce the highest prescient yield for the the recrystallization of ammonium bromide?
a) a solution containing 120 g NH4Br in 100 g H20 at 80 C is cooled to 20 C
b) a solution containing 95 g NH4Br in 175 g H20 at 80 C is cooled to 0 C
c) a solution containing 115 g NH4Br in 200 g H20 at 80 C is cooled to 0 C
d) a solution containing 100 g NH4Br in 100 g H20 at 80 C is cooled to 20 C
e) a solution containing 50.5 NH4Br in 100 g H20 at 20 C is cooled to 0 C
0.0354 mg/mL at 1 atm and you want to convert this M (moles/L) at 0.00036 atm.
0.0354 mg/mL*0.00036 = 1.27E-5 mg CO/mL.
1.27 mg CO/mL x (1000 mL/1L) x (1 g/1000 mg) x (1 mol CO/28 g CO) = 4.5E-7 moles/L = 4.5E-7 M
You need to go through these and calculate each one.
A. You have 120 g. The solubility is 125 g/100 g H2O; therefore, all of it will dissolve at 80 C. If cooled to 20 C, all but 76.4 will crystallize. How much is that? 120-76.4 = 43.6g recovered.
B. You have 95 g in 175g H2O. The solubility is 125 g/100 g H2O; therefore, all of this will dissolve at 80 C. The solubility at 0 C the solutility is 60.5 g/100. You have 175 mL; therefore, that 175 at 0 C will hold back 60.5 x 175 gH2O/100 gH2) = about 106 g. How much of the original 95g will you recover? none (< 0).
C. Do all of them like this (A and B are examples of the different types you may have). You will find that A will do the best for you. I found < 0 recovered from C, about 40 g for D and < 0 for E.
I understand question 2. But for question one, why did you multiply by (1000 mL/1L) x (1 g/1000 mg) in that way? I understand you did it to get the units right, but why was the 1000 mL on top for one and 1 g on top for another?
To convert to the right units should be answer enough but let me go the long way around. Suppose we wish to change 1.5 feet to inches.
1.5 feet x factor = ? inches.
We need two things.
1. We need a conversion factor
2. We need to place the factor correctly.
#1 first. What is the factor? We know there are 12 inches in a foot; therefore, the two conversion factors we have (there are ALWAYS two factors; i.e., one is inches to feet and the other is feet to inches) are
(12 inches/1 foot) AND (1 foot/12 inches).
#2. How do we place the factor properly? We place the factor so that the unit we don't want to keep cancels but the unit we want to keep stays.
Here's how it works.
1.5 feet x (12 inches/1 foot) = 18 inches.
Notice that the feet unit (the one we want to get rid of) cancels since one is in the numerator and the other in the denominator. Note also that we want to keep the inches unit and that one we leave untouched. What if we thought the other way is correct? Let's try it.
1.5 feet x (1 foot/12 inches) = 0.125 ft2/inch. We wanted inches and feet2/inch isn't it.
So the reason 1 placed the units as I did is that one factor was (1g/1000 mg) and the other was (1000mL/1L).
The (1g/1000 mg) was converting mg to g) and the other converted mL to L. I hope this helps.