Question

The true absorbance of a sample is 1.000, but the mono chromator passes 1.0% stray light. Add the light coming through the sample to the stray light to find the apparent transmittance of the sample. Convert this answer back into absorbance and find the relative error in the calculated concentration of the sample.

Short answer

The absorbance is 0.9629.

Step-by-step solution

Calculate the apparent Transmittence:

Given,

The true absorbance of a sample is 1.000, but 1.0 % stray light reaches the detector.

The apparent transmittance can be calculated as

True transmittance$={10}^{-1.000}=0.100$ with 1.0 % stray light, the apparent transmittance is

$$\begin{gathered} \frac{\mathrm{P}+\mathrm{S}}{{\mathrm{P}}_0+\mathrm{S}}=\frac{0.100+0.010}{1+0.010} \\ =\frac{0.11}{1.01} \\ =0.1089 \end{gathered}$$

Step:2  Apparent absorbence:

The apparent absorbance can be calculated using -log T

-log0.1089

= 0.9629