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Molar absorptivity is a property of a substance which determines the fraction of light absorbed by it at a particular wavelength. Absorbance is the fraction of light absorbed.

$A=\mathrm{Î\mu }cb$

Where,

$$\begin{gathered} \mathrm{Î\mu }=\mathrm{molarabsorptivitycoefficient} \\ \mathrm{b}=\mathrm{lengthofcuvet} \\ \mathrm{c}=\mathrm{concentrationofsolution} \\ \mathrm{A}=\mathrm{absorbance} \end{gathered}$$

Theconcentration of the solution is$3.96×{10}^{-4}\mathrm{M}$

Absorbance (A)$=0.624$

Wavelength$\left(\mathrm{λ}\right)=238\mathrm{nm}$

Length of cuvette$\left(\mathrm{b}\right)=1.00\mathrm{cm}$

Absorbance for blank solution is at$238\mathrm{nm}=0.029$

Therefore, the molar absorptivity is calculated below.

$$\begin{gathered} \mathrm{Î\mu }=\frac{\mathrm{A}}{\mathrm{bc}}=\frac{\begin{array}{c}\mathrm{A}=\mathrm{Î\mu ³¦²ú}\\ 0.624-0.029\end{array}}{(196×{10}^4{\mathrm{M}}^{-1}){\mathrm{cm}}^{-1}\left)\right(1.000\mathrm{cm})} \\ =1.50×{10}^3{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$

Absorbance (A)$=0.375$

Wavelength $\left(\mathrm{λ}\right)=238\mathrm{nm}$

Length of cuvette$\left(\mathrm{b}\right)=1.00\mathrm{cm}$

Absorbance for blank solution is at $238nm=0.029$

Molar absorptivity of A$\left(\mathrm{Î\mu }\right)=1.50×{10}^{-3}{\mathrm{M}}^{-}{\mathrm{cm}}^{-}$

Concentration of unknown compound A is calculated below.

$$\begin{gathered} \mathrm{c}=\frac{\mathrm{A}}{\mathrm{Î\mu ²ú}}=\frac{0.375-10.029}{1.50×{10}^3{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1}\left(1000\mathrm{cm}\right)} \\ =2.31×{10}^{-4}\mathrm{M} \end{gathered}$$

Initial volume$=2.00\mathrm{mL}$

Final volume = 25.00ml

Absorbance (A) =0.743

Wavelength$\left(\mathrm{λ}\right)=238\mathrm{nm}$

Length of cuvette$\left(\mathrm{b}\right)=1.00\mathrm{cm}$

Absorbance for blank solution is at$238\mathrm{nm}=0.029$

Molar absorptivity of A $\left(\mathrm{Î\mu }\right)=1.50×{10}^{-3}{\mathrm{M}}^{-}{\mathrm{cm}}^{-}$

The concentration of A is given below.

$$\begin{gathered} \mathrm{c}=(\mathrm{dilution}\mathrm{factor})\frac{\mathrm{A}}{\mathrm{Î\mu ²ú}} \\ \mathrm{c}=\left(\frac{25.00\mathrm{ml}}{200\mathrm{ml}}\right)\frac{0.733-0.009}{\left(1.50×{10}^3{\mathrm{M}}^{-1}{\mathrm{tm}}^{-1}\right)\left(1.000\mathrm{m}\right)} \\ =5.87×{10}^{-3}\mathrm{M}. \end{gathered}$$