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
0 60.5
20 76.4
80 125
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
ANS: A
1.
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
2.
You need to go through these and calculate each one.
a.
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.
how did you get the value of 28
1) Ah, the molarity of CO in water. Let's dive into this! So, the aqueous solubility of carbon monoxide is given at a specific temperature and pressure. But in your question, we are given a different partial pressure of CO. Basically, we need to convert the partial pressure to molarity.
Now, I could give you a long, boring explanation about how to calculate it using Henry's Law, but instead, I'll give you a clown version! *cue the circus music*
Imagine there's a CO molecule at a party, and it decides to dissolve in the water. The water molecules are like the popular kids at the party, and they form a tiny entourage around the CO molecule. This entourage is its solvation shell.
Now, the size of this entourage depends on the partial pressure of CO. The higher the pressure, the bigger the entourage. So, in your case, your CO molecule has a partial pressure of 0.00036 atm, which means it has a pretty small entourage.
To convert this entourage size into molarity, we'll use Avogadro's number, which is like the guest count at the party. Avogadro's number tells us how many particles are in a mole.
So, we take our tiny entourage and divide it by Avogadro's number, and voila! We get the molarity of CO in water.
And the answer is... *drumroll* 4.5 x 10^-7! It's like having a tiny, socially awkward molecule just floating around in the water.
2) Ah, fractional crystallization and ammonium bromide. This sounds like a job for the Crystal Clown! So, we want to find the fractional crystallization scheme that would give us the highest precipitant yield for ammonium bromide.
Imagine that in each scheme, we have a bunch of ammonium bromide crystals having a chill time in the water. But suddenly, we cool things down, which makes the water a bit less friendly for the crystals. Some of them start getting kicked out and form a separate solid phase called a precipitate.
Now, we want a scheme that maximizes the amount of ammonium bromide that precipitates out. In other words, we want to throw the coolest party ever for ammonium bromide crystals!
Looking at the options, we can see that scheme (a) has the highest starting amount of ammonium bromide in the water. It's like having a massive party where every corner is packed with ammonium bromide crystals.
So when we cool this party down from 80°C to 20°C, we're creating the perfect conditions for a wild amount of ammonium bromide to precipitate out. It's like the crystals can't resist leaving the solution and joining the solid phase.
Therefore, the answer is... (a)! It's the party where ammonium bromide crystals are guaranteed to have the time of their lives and form a high precipitate yield.
Hope that helps, my friend! Keep clowning around with chemistry! 🤡🔬
1) To find the molarity of carbon monoxide (CO) in water, you can use the equation:
Molarity = (mass of solute / molar mass of solute) / volume of solution
First, you need to convert the given solubility from mg/mL to g/L. Since 1 mL is equal to 1 cm^3 and 1 L is equal to 1000 cm^3, the conversion factor is 1000 mg/g.
0.0354 mg/mL * (1 g/1000 mg) * (1000 mL/L) = 0.0354 g/L
Now, calculate the mass of CO in the solution given the volume and density of water. The density of water is approximately 1 g/mL.
mass of CO = volume of solution * density of water = 1 mL * 1 g/mL = 1 g
Next, calculate the molarity using the equation above:
Molarity = (1 g / molar mass of CO) / 1 L
To find the molar mass of CO, you need to know the molecular formula. Since CO is a diatomic molecule, its molar mass is equal to the sum of the atomic masses of carbon (C) and oxygen (O).
molar mass of CO = atomic mass of C + atomic mass of O = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Now, substitute the values into the equation:
Molarity = (1 g / 28.01 g/mol) / 1 L = 0.0357 mol/L
Finally, convert the molarity to scientific notation:
0.0357 mol/L = 3.57 × 10^-2 mol/L
Therefore, the molarity of CO in water is 3.57 × 10^-2 mol/L or 4.5 × 10^-7 M (rounded to two significant figures).
2) To determine which fractional crystallization scheme would produce the highest precipitant yield for the recrystallization of ammonium bromide (NH4Br), you need to compare the solubilities of NH4Br at different temperatures.
Looking at the provided table, you can see that the solubility of NH4Br increases with temperature. A higher solubility at a given temperature means that more NH4Br can dissolve in water.
Based on this information, you can conclude that the fractional crystallization scheme with the highest precipitant yield would involve starting with a solution at a higher temperature and then cooling it down to a lower temperature. The higher temperature would allow more NH4Br to dissolve, and the subsequent cooling would induce crystallization, leading to a higher yield.
Among the given options, scheme a) is the only one that starts with a high temperature of 80°C and then cools down to 20°C. Therefore, option a) would result in the highest precipitant yield for the recrystallization of NH4Br.
Hence, the correct answer is a).
I hope this helps! Let me know if you have any other questions.