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# dilution factor calculator

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Question:Consider the following experiment in which you wish to determine the molarity of the concentrated HCl in your new bottle of reagent. You carefully measure out 100.0 mL of the concentrated acid and dilute it to 1.000 L (in a one-liter volumetric flask). You then use a 5.00 mL volumetric pipette to remove 7 aliquots for titration with 0.1068 M sodium hydroxide solution. the mean valie of sodium hydroxide was 59.9 mL, or 0.0599L, then i multipled by 0.1068 to find the moles of NaOH. because its a one to one ratio, then the moles of the acid is the same (~0.00634732 mol)...now i dont know wat my next step should be..i know i have to take the dilution into consideration..

Answers:Since you already have the moles of acid, just divide by the volume in liters. You diluted it by a factor of ten (from 100 to 1000 mL), so, if you like, just divide by 0.0005 instead of 0.005 to account for it and you're done.

Question:I started with 1g of a herb having its metal content analysed by AAS. Now, I have multiplied the result that I got by the dilution factor to give me xmg/L or ppm. But I need to determine the amount of metal present in the 1g of herb, how would I go about doing this? I seriously cannot wrap my mind around this..

Answers:Did you extract the entire 1 g of herb??? If so then all of the metal from the 1 g of herb is in the original volume of solution that you started with. So apply your x mg/L to the volume of your extract. eg you had 50.0 mL of extract, all the metal from 1 g is in this 50.0 mL. It has conc. of 0.2 mg/L 50.0 mL = 0.0500 L So amount in 1 g = 0.2 mg/L x 0.0500 L = 0.01 mg in 1 g herb = 0.01 mg/g

Question:"If you mix 10 L of unknown solution with 90 L of water, then took 5 L of the mixture and mixed it with 3 mL of Bradford reagent, then what is the total dilution factor?" This was a question on my biochemistry lab exam. There is two dilution factors since there are two dilutions. The first dilution factor is: (90 L water + 10 L of unknown solution/10 L of unknown solution)= 10. My professor marked that answer right. However, I had trouble with the second dilution. The second dilution factor is: (5 L of the mixture + 3000 L of Bradford reagent/ 5 L of the mixture)= 601. He marked this wrong. So the total dilution factor is: 10 X 601= 6010. I knew this was wrong since this number is way too high. So what is the right answer? Thanks.

Answers:First dilution is 10 fold. Second dilution is 600. Total dilution is 6000. Usually, however, when using the Bradford assay for protein, the second dilution is not taken into account because all standards, and unknowns, etc., are run this way. It isn't considered a dilution, but rather the addition of assay reagent.

Question:Also, I have calcium gluconate, and then i add 5 ml of MgSO4 and 10ml of ammonium buffer, are those additives counted as part of the calcium gluconate? or part of the disodium edentate that i use for titrating?

Answers:You should dilute the calcium gluconate (determine the dilution factor) befrore you add titrants, buffers etc. Unless these add a significant volume. Shanti