Determination of Crude Protein Content (Part 2)_Chemical Preparation (AOAC 2001.11) 

Tarfaoui Louiza, you are most welcome. Stay with us.

Just dissolve 1g in 100ml ethanol

Anees Hayat, Thanks for your comment. Please find the answer of your question given below: HCL - 37% v/v. Specific gravity: 1.19 g/ml 37ml/100 ml or 370 ml/1000 ml x 1.19 g/ml = 440.3 g/L HCL Molecular weight = 36.5 Molarity: 440.3 grams /36.5 grams = 12.06 M or ~12M ==================== Compounding 1 liter of 0.1N Solution ==================== M1V1 = M2V2 (0.1)(1000) = (12) (x) x = (0.1) (1000) / 12 x = 8.3 ml ***Therefore add 0.83 ml of 37% HCL to 100ml of Distilled Water to create a 0.1N HCL solution.

@@MicroChemsExperiments thank u so much

@@aneeshayat2178 you are most welcome

Sorry we can't suggest a text book. Plzz take help from google. But you can purchage AOAC manual book.

Please learn about burette at your lab

Thanks

It is a milk product. So use 6.38 as the protein factor

@@MicroChemsExperiments Alright, Thank you!

You're most welcome

Raju Ahmmed, Thank you for your comment. Unfortunately we dont have any tutorial of Biuret Method. In future, we will develop the method. Stay with us.

No. You have to use HCl.

36g is the actual value. Prepare catalyst exactly as done in this video

@@MicroChemsExperiments Okay thank you so much.. I'm doing this experiment exactly the same.. It's really helpful..

No. You should use Selenium dioxide

Ok

No. You must need three reagents to be used.

Different analysts modified this method in different ways. Thats why there is slight difference in the catalyst preparation

Methyl red indicator is used during titration after distillation in the part 1

@@MicroChemsExperiments Thank you sir

Slightly modified from AOAC method but internally validated in our lab which gives accurate results

@@MicroChemsExperiments i didn't get it

No

Stock bottle of 37% HCL. Determination of molarity of 37% HCL V/V 37 ml of solute/100 ml of solution. HCL - 37% v/v. Specific gravity: 1.19 g/ml 37ml/100 ml or 370 ml/1000 ml x 1.19 g/ml = 440.3 g/L HCL Molecular weight = 36.5 Molarity: 440.3 grams /36.5 grams = 12.06 M or ~12M ==================== Compounding 100ml of 0.1N Solution ==================== M1V1 = M2V2 (0.1)(1000) = (12) (x) x = (0.1) (100) / 12 x = 0.83 ml Therefore add 0.83 ml of 37% HCL to 100ml of DW to create a 0.1N HCL solution.

Yes.

It means initial burette reading. Plzz follow the mark on burette scale.

Sir I do experiment as Per your Procedure I got reading 24.ml But when I take blank reading it show only o.1 ml reading. So reading difference increases. how do its percentage as per your formula

It was not clear to me. Plzz write and explain in details. I didn't do any blank titration. So where did you find blank reading?

Sir, I take 2 gm Sample+2 gm catalyst mixture+20 ml H2SO4 +2 hr digestion After that I prepare 100 ml Sample Solution.10 ml Sample is use from this Solution in kj flask+50ml 40% NaOH+50 ml distill water from this 100 ml distillate were titrate with .1N HCl I got reading :24 ml After that my Problem is not Solve.Percentage get high.

@@ranganathgolhar1063 Your result is invalid. Burette reading 24 is not possible. Did you collect your distallate in 30ml of 4% boric acid solution? Which sample did you test?

Part three is not needed. For protein analysis, Part 1 & 2 is enough

Yes. It is