91Ó°ÊÓ

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.

$\mathrm{A}=\mathrm{Î\mu ³¦²ú}$

Where,

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

The mass of${\mathrm{NH}}_4\mathrm{Cl}\mathrm{is}1.00×{10}^{-2}\mathrm{g}.$

The concentration of${\mathrm{NH}}_4\mathrm{Cl}$is calculated below.

$\frac{{\displaystyle \frac{1.00×{10}^{-2}\mathrm{g}}{53.491\mathrm{g}/\mathrm{mol}}}}{1.00\mathrm{L}}=1.869×{10}^{-4}\mathrm{M}$

The concentration of coloured solution${\mathrm{NH}}_4\mathrm{Cl}$

$\frac{10\mathrm{mL}}{50\mathrm{mL}}(1.869×{10}^{-4}\mathrm{M})=3.739×{10}^{-5}\mathrm{M}$

The molar absorptivity :

$$\begin{gathered} \mathrm{Î\mu }=\frac{\mathrm{A}}{\mathrm{bc}}=\frac{(0.308-0.140)}{(1.000)(3.739×{10}^5\mathrm{M})} \\ =4.493×{10}^3{\mathrm{M}}^{-1}{\mathrm{cm}}^{-1} \end{gathered}$$

Mass of${\mathrm{NH}}_4\mathrm{Cl}\mathrm{is}1.00×{10}^{-2}\mathrm{g}.$

Concentration of reference solution $1.869×{10}^{-4}\mathrm{M}$

Mass protein value 4.37mg

The calculation concentration ${\mathrm{NH}}_3$of the unknown solution is done below.

$$\begin{gathered} \frac{Absorbanceofunknown}{Absorbanceofreference}=\frac{concentrationofunknown}{concentrationofreference} \\ ⇒\frac{0.592-0.140}{0.3080.140}=\frac{Concentrationofunknown}{1.869×{10}^{-4}} \\ ⇒concentrationofunknown=\frac{0.452}{0.168}(1.869×{10}^{-4}=5.028×{10}^{-4}M{N}_3. \end{gathered}$$

The amount of mass of Nitrogen present in 100 ml solution is given as:

$5.028×{10}^{-5}\mathrm{mol}\mathrm{of}\mathrm{N}×14.0067\mathrm{g}=0.7043\mathrm{mg}\mathrm{of}\mathrm{N}$

Calculating the weight percent of nitrogen in the unknown solution:

$$\begin{gathered} wt\%ofN=\frac{100×0.7043mg}{4.37mg} \\ =16.1\% \end{gathered}$$