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To prepare the Iest solution, 50 mL vo umotric flask add 50 mL 100 mgi Fes+ solution from Ihe 10 mL buretto. With the provided abelled, glass pipette, add 00 mL of...

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To prepare the Iest solution, 50 mL vo umotric flask add 50 mL 100 mgi Fes+ solution from Ihe 10 mL buretto. With the provided abelled, glass pipette, add 00 mL of 2000 M KSCN solutlon: Miix well and add mL; using auto-dispenser of Mnitric acid. Make the solution up to Ihe 50 mL mark , slopper Ine flask and mlx well. Note that the concenfrarion [hocyanale colutio n the fest solulion only comdared Kith for the standards; comiete converrian tno comolox doos not occur Calculate the concentrations

To prepare the Iest solution, 50 mL vo umotric flask add 50 mL 100 mgi Fes+ solution from Ihe 10 mL buretto. With the provided abelled, glass pipette, add 00 mL of 2000 M KSCN solutlon: Miix well and add mL; using auto-dispenser of Mnitric acid. Make the solution up to Ihe 50 mL mark , slopper Ine flask and mlx well. Note that the concenfrarion [hocyanale colutio n the fest solulion only comdared Kith for the standards; comiete converrian tno comolox doos not occur Calculate the concentrations of Fe * and Fe(NCSY- your propared solutions and wnte them in the relevant column the table on tho noxt page. Calculation of [Fo" ] concontration the standard solutions: Start by vorking out the molar concenbalion oltho 100 mG/L ("'stock ) Ircn solution You used make Incm Up_ mgIL Moll [Fe" ] stock solulion Now calculate Ihe concentraton (n M) = pacn dilution using the equation: M,V MzVz Which can rearrangud produce the equation for the quanbty we want: Mz(Vz / Va) whare: is Ine concentrallon 0f Foj" In the fInal solution, Va [s the volurne of the Fel final solumion tne volume of tho volumetnic flask; this case 50&L thu volume Ihe sock solution you added the flask (in mL) molanIFe concentraton Ihe stock soluton from YOUt calculatlon above Anannernin crocenon Lade Irie Odulion ratlo long e3 ;oJ use Urr corllertiona Iud boln volumus aucn Fcu Klarunno concuntratiom Tnaetandurusolubons: Calculatlon = [Fe(Ncs)" ] havu added excust thlocyanate. tume (hu complo A4CluSt Llmtosllcomdelely Iherprdicleidu lormallcn #quilibnum Ine Fe NCSk' tha elandard tolulions naranand concuntration Reumud oquivatent t0 thu concaniralion



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Purification of proteins for use as biopharmaceuticals is often accomplished by ion exchange chromatography, in which a process fluid passes through a column packed with small resin beads whose ionic surface charge causes them to adsorb some stream components more strongly than others. An ion-exchange run takes place in two steps: (1) the load step, in which the process stream flows through the column and the target protein (the product) and some undesired impurities are adsorbed onto the resin; and (2) the elution step, during which another fluid passes through the column and desorbs the impurities and the protein from the resin. The elution fluid consists of an aqueous solution of a solute known as Tris diluted with an $\mathrm{NaCl}$ solution, with the NaCl-to-Tris ratio starting at 0 and steadily increasing with time. The impurities desorb into the fluid when the NaCl concentration is low, and the effluent is collected in a waste vessel. As the NaCl concentration increases, the target protein desorbs. When analysis of the effluent reveals the presence of the target protein, the flow is switched to the product collection vessel, and the effluent is collected until no more product is detected in the effluent. The collected product is then subjected to additional process steps to further isolate the protein, and the column is cleaned for reuse. Consider an elution step in which solutions of 1 M $\mathrm{NaCl}$ (solution A) and 50 mM Tris (solution B) are mixed and fed to a loaded ion-exchange column. The system is programmed to keep the total volumetric flow rate $\left(\dot{V}_{\Lambda}+\dot{V}_{B}\right)$ into the column constant at 120 Lh while linearly increasing the volume fraction of solution A in the feed from $0 \%$ to $20 \%$ over a period of 33.6 minutes, at which point the elution is declared to be complete. The flowchart is shown below:
(a) Calculate $V_{\mathrm{t}}(\mathrm{L}),$ the total amount of solution fed to the column. (b) Derive an equation for the volumetric flow rate of solution $A, \dot{V}_{A}(t),$ assuming that the densities of both fluids are the same. Use the calculated value to determine calculate $V_{\mathrm{A}}(\mathrm{L}),$ the total volume of that solution fed to the column, and $m_{\mathrm{A}}(\mathrm{g} \mathrm{NaCl})$, the total mass of $\mathrm{NaCl}$ fed. Then determine $\dot{V}_{B}(t)$ and $V_{\mathrm{B}}(\mathrm{L})$ and $n_{\mathrm{B}}(\text { mol } \text { Tris) }, \text { the total volume of solution } \mathrm{B}$ and total moles of Tris fed, respectively. (Hint: Once you've done the calculations for solution A, those for B should be trivial.) (c) Suppose in one run product is detected in the effluent at the same time impurities are detected $-$ that is, product protein starts desorbing earlier than in previous runs. List up to five possible causes of the problem.

Take a has led to problems. 17. We need to put the television plans for mass fraction against solution density. We need to do this problem in exile, however, for illustration, let's see how it looks. So the first point will be in the densities they want 988 yeah, and the second density is around 0.99 And to the mass ratio is those 21 for the 2nd 10.992 It is a little bit more than 0.15 and that's how the other three points would be at. And we will get a straight line, and then straight line will have this situation. Why? Equal to 545.5 x minus 539.3 So here Snoopy is 545.5, and the intercept will be at negative 5 39.3 and the R C Square is 0.9992 Therefore, the equation for the straight line thought is, master, are you can do 545.5 underplayed to that density Negative 539.3 Next. When he goes to the mass florid, it is given that the intensity the density is 0.994 Grandpa sentiment Turkey for a solution. Therefore, let's calculate what is the mass ratio Mass ratio will be 545.5, $25.0.994 Grandpa Centimeter Cube minus 539.3 which gives us the mass ratio is 303.197 gram of my solution 100 runoff water. Therefore, there is 3.197 grandma find solution in 100 gram of water and then my story, therefore, is volume fluid quantified by density multiplied wider mass ratio. Holding through it is even as 1 50 liters by our let's convert it to centimeter to 10 centimeters, divided by one leader. The density is you're going to 994 grand person to interview, and the ratio is 3.1 97 g of a solution divided by 100 a gram of water. Let's get it to G one kg, divided by 101,000 g. Therefore, we get the mass for Eddie's 4.76 kids, Your solution. Bar power. Next, we need to excellent or the mass already is too high or too low the mass slaughter that were calculated If it is high to hire two, though now in this problem, we assume that the a solution density increases with decreasing temperature. We asked them that the solution density yeah, in process right, represent temperature. Therefore, the solution in city at 47 degrees Celsius is more than 0 29 4 grand park sentiment ridicule. Therefore, the mask, already calculated in part B, is too low.

In question 1 11. It talks about how external pressure must be applied more than the osmotic pressure in order to allow solvent from a hi concentrated solution on one side of a semi permeable membrane to send its water molecules through the membrane to the other side. That might be pure water. This osmotic pressure obeys the laws similar to the ideal gas law where P. B equals Nrt. It then shows you that if we replace P with osmotic pressure that we get an over V. Which is related to more clarity. So we have osmotic pressure being equal to polarity, multiplied by RT. With this equation, it asks you to determine the osmotic pressure at 25 degrees Celsius of a 250.20 molar sucrose solution. Well, sucrose only separates, well, it doesn't separate into anything one unit of sucrose creates uh one mole of sucrose. Well, one molecule of sucrose doesn't separate like ionic compounds do. So it's polarity is it's osmolarity. So when using the equation, osmotic pressure will be equal to the straight up polarity of the sucrose solution. 0.20 Multiplied by are multiplied by T. The r value is 0.8 to 06 Leader atmospheres per kelvin mole. So we'll end up getting pressure in units of atmospheres when using this equation. T however, needs to have units of kelvin because our our value has units of kelvin innit leader atmospheres per kelvin mole. Therefore will convert the 25 degrees Celsius. That is provided into kelvin temperature by adding to 73 and we get an osmotic pressure for the solution of 0.49 atmospheres. For part B. It talks about seawater containing 34 g of salts. For every leader of solution that wants you to assume that the salute consists entirely of sodium chloride and to calculate the osmotic pressure of seawater 25 degrees Celsius. We'll do the same thing as we did before. Osmotic pressure will be equal to the polarity multiplied by are multiplied by T. But this needs to be the osmolarity referring to the total polarity of all the salute ions in solution. So if we have 3.4 g of sodium chloride per liter, we can convert the gram sodium chloride into molds sodium chloride by dividing by the molar mass sodium chloride. So we will then have the grams will cancel will have units of mold sodium chloride per liter sodium chloride. That will be the more clarity sodium chloride. But the osmolarity of the solution because sodium chloride separates completely into sodium ions and chloride and ions will be two times that. So this whole thing right here will be the osmolarity or the end value. Well then multiply that by R. Value. And then the temperature again 25 degrees Celsius plus 2 73 to get our kelvin temperature and we get 285 atmospheres for part C. It states that the average osmotic pressure of blood is 7.7 atmospheres at 25 degrees Celsius. So what concentration of glucose will be isotonic with blood? Well, we'll use the same osmotic pressure equation but now we'll be solving for the osmolarity of a glucose solution. Knowing the pressure, if the pressure is 7.7 atmospheres, that'll be set equal to the osmolarity of the glucose solution. Multiplied by the R value. Same as always 0.8 to 06 Multiplied by the kelvin temperature. Which will be the 25 plus 2 73 or 2 98 kelvin rearrangement of this equation gives us an osmolarity of 315 Mueller for glucose because glucose is not separate into ions. It's osmolarity. Is it straight up polarity now for part D. Why says I'm is an enzyme that breaks bacterial cell walls. The solution that contains 0.150 g of the enzyme in 212 mL of solution has an osmotic pressure a 2120.953 tour at 25 degrees Celsius. And for this question, it asks you to calculate the molar mass of the list design. Well, if we go back up here, you'll notice where we put in the molar mass, we put in the molar mass right here. So if we know the pressure and we don't know the molar mass, then this is where our unknown variable will be found in our calculation. It'll look something like this. We'll take the pressure and tour and converted into atmospheres by dividing by 7 60. This will then be osmotic pressure and atmospheres. We'll set that equal to the mass that we have multiplied by one over the molar mass as we did up here. One over the molar mass. Well then divide that whole thing by the leaders of the solution. The two 10 mL can be converted into leaders by dividing by 1000. So now we have this right here. If we knew molar mass would give us the polarity of the license. I'm solution because the licensee does not separate into ions which was not necessarily obvious in the question. But if it's not a common ionic compound, you can assume that the more clarity is the osmolarity and it does not separate this whole thing right here then is the polarity. We will multiply that by the R. Value and then by the T. Value. And all we have now is an algebraic equation for which we can solve for the molar mass. So I'll save you some space here and not take you through all the algebra. But the molar mass ends up becoming 13,900 g per mole. Then for part E. Osmotic pressure of acquis solution of a certain protein was measured in order to determine the protein smaller mass. The solution contained 3.50 mg of dissolved protein in five mL. The osmotic pressure at 25 degrees Celsius was 1.542 Or what is the molar mass. So this will be another calculation very similar to what we did up here. We'll take the tour converted into atmospheres by dividing by 7 16. We'll set that equal to the mass, multiplied by one over the unknown molar mass in order to get moles. Well then divide the moles by the volume. In order to get more clarity. five mL can be converted into leaders by dividing by 1000. Mhm. This whole thing here then is the polarity. Well multiply that by there are value 0.8 to 06 Multiply that by the temperature. And then again, all we have left is an algebraic expression for which we will solve the molar mass. So without going through all the algebra. Hopefully, you know, enough to get through the algebra, you just need to isolate this essentially what we're doing is we're gonna multiply both sides by mm mm is going to come over here and then we'll divide both sides by this number right here. When we do that, we have mm equal to all of this over here. Except that part with this being divided on both sides. And we get 8910 g per mole for answer


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