Question

I'm a might stuck on some calculations for my lab report on Spectrophotometric Determination of Protein.

I'm looking to calculate the concentration of the first standard.

Sample

Vol. of Protein Std Solution

Vol. of BCA reagent

Volumeof DI water

Total Vol.

Vol of Unknown

Blank	0	3.00	1.000	4.00	0
Standard 1	0.100	3.00	0.900	4.00	0
Standard 2	0.200	3.00	0.800	4.00	0
Standard 3	0.300	3.00	0.700	4.00	0
Standard 4	0.400	3.00	0.600	4.00	0
Standard 5	0.500	3.00	0.500	4.00	0
Unknown	0	3.00	0.900	4.00	0.100
Unknown	0	3.00	0.900	4.00	0.100
Unknown	0	3.00	0.900	4.00	0.100

The concentration (mg/mL) of the protein standard solution is 1.0074 mg/mL (or 503.7mg/500.00mL)

So I think the concentration of Standard 1 is: (The mass of the protein standard solution * volume of protein standard solution added)/ the total volume

So Standard 1 is: (1.0074 mg/mL * 0.100 mL)/ 4.00 mL = 0.025185 mg/mL.

I just need to know if I calculated this correctly. Thank you!

Answer 1

Page No. Date concentration of stock solution = 1.0074 mg/ml Total volume prepared = 4 ml Volume of protein Standard Solution = 0.1 ml Formula: Niv = N2V21 Concentration of standard Volume of std. protein X protein = mila. I I cance of diluted soin. - of protein M = 1.00 74 mg/mL, V = N₂ = ? , V2 = 4 m/ Volume of protein solution prepared 0.1 ml Here of i Niv, = N2V2 1.0074 X 0,1 = Ne X 4 . No 1.0074 Xool N₂ = 0.025185 mg/ml answer is You perform correct. good carry calculation. Your on ...