Question

What mass of protein gives a corrected absorbance of $\mathbf{0}\mathbf{.}\mathbf{250}$?

Short answer

The mass of protein giving a corrected absorbance of $0.250\mathrm{is}15.0\mathrm{\mu g}$.

Step-by-step solution

Concept Used

Linear Calibration Curve:

The equation of the linear calibration line is:

$\mathrm{y}\left(\pm {\mathrm{s}}_{\mathrm{y}}\right)=\left[\mathrm{m}\left(\pm {\mathrm{u}}_{\mathrm{m}}\right)\right]\mathrm{x}+\left[\mathrm{b}\left(={\mathrm{u}}_{\mathrm{b}}\right)\right]$

where,

$\mathrm{y}$is the corrected absorbance.

Corrected Absorbance = Observed Absorbance - Blank Absorbance

Calculation of Mass of Protein

The equation of the linear calibration line is:

$$\begin{gathered} \mathrm{absorbance}=\mathrm{m}\mathrm{x}\left(\mathrm{\mu g}\mathrm{of}\mathrm{protein}\right)+\mathrm{b} \\ \mathrm{y}=\left(0.00163\right)\left(\mathrm{\mu g}\mathrm{of}\mathrm{protein}\right)+0.{004}_7 \end{gathered}$$

The numbers in the subscript denote insignificant figures.

The given corrected absorbance $\left(\mathrm{y}\right)=0.250$

The mass of protein is:

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

Mass of Protein

$$\begin{gathered} =\frac{0.250-0.0047}{0.01630} \\ =15.05 \\ =15.0_{\mathrm{\mu g}}\left(\mathrm{rounded}\mathrm{off}\right) \end{gathered}$$

Therefore, the mass of protein resulting in a corrected absorbance of $0.250\mathrm{is}15.0\mathrm{\mu g}$