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Chapter 0: Q4-9 TY (page 1)

What mass of protein gives a corrected absorbance of $0.250$ ?

Short Answer

Expert verified

The mass of protein giving a corrected absorbance of $0.250 is 15.0 μg$ .

Step by step solution

01

Concept Used

Linear Calibration Curve:

The equation of the linear calibration line is:

$y (\pm s_{y}) = [m (\pm u_{m})] x + [b (= u_{b})]$

where,

$y$ is the corrected absorbance.

Corrected Absorbance = Observed Absorbance - Blank Absorbance

02

Calculation of Mass of Protein

The equation of the linear calibration line is:

$absorbance = m x (μg of protein) + b y = (0.00163) (μg of protein) + 0 . 004_{7}$

The numbers in the subscript denote insignificant figures.

The given corrected absorbance $(y) = 0.250$

The mass of protein is:

$\frac{= corrected absorbance - 0.0047}{0.01630}$

Mass of Protein

$= \frac{0.250 - 0.0047}{0.01630} = 15.05 = 15 . 0_{μg} (rounded off)$

Therefore, the mass of protein resulting in a corrected absorbance of $0.250 is 15.0 μg$

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Most popular questions from this chapter

A titration curve for Na2CO3 titrated with HCl is shownhere. Suppose that both phenolphthalein and bromocresol green arepresents in the titration solution. State what colors you expect to

observe at the following volumes of added HCl:

2 mL; (b) 10 mL; (c) 19 mL.

A solid mixture weighing 0.05485 gcontained only ferrous ammonium sulphate and ferrous chloride. The sample was dissolved in $1 {MH}_{2} {SO}_{4}$ , and the ${Fe}^{2 +}$ required 13.39 mL oflocalid="1663577431957" $0.01234 {MCe}^{4 +}$ for complete oxidation to localid="1663577453691" ${Fe}^{3 +}$ . Calculate the weight percent of Cl in the original sample. If you need refreshing, see Section 7-2 for an example of a titration of a mixture.

localid="1663574328546" ${FeSO}_{4} \times {({NH}_{4})}_{2} {SO}_{4} \times 6 H_{2} O {FeCI}_{2} \times 6 H_{2} O$

Ferrous ammonium sulfate Ferrous chloride

FM 392.13

FM 234.84

Effect of concentration in the titration of weak acid with strong base.Use your spreadsheet from Problem 11-66 to prepare a family of titration curves for pKa = 6, with the following combinations of concentrations: (a) C_a = 20 mM, C_b = 100 mM; (b) C_a = 2 mM, C_b = 10 mM; (c) C_a = 0.2 mM, C_b = 1 mM.

Find the minimum detectable concentration if the average of the blanks is $1 . 0_{5} n A$ and $0 . 6_{3} n A$ .

Titration of EDTA with metal ion. Use Equation 12-12 to reproduce the results of Exercise 12-C

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