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Amount of nitrogen in protein =16.2 wt%

Amount of protein solution =0.5 mL

Amount of HCl in which liberated NH₃ was distilled = 10mL of 0.0214 M HCl

NaOH required for titrating unreacted HCl =3.00 mL of 0.0198 M NaOH

This problem can be solved by using the method of Kjeldahl analysis. This method is widely used in determining the presence of nitrogen in different organic substances.

The initial number of moles of HCl in the receiver

$$\begin{gathered} =\left(10×0.0214\right)mmol \\ =0.214mmol \end{gathered}$$

Moles of NaOH required for titrating unreacted HCl

$$\begin{gathered} =3×0.0198mmol \\ =0.0594mmol \end{gathered}$$

Therefore, the number of moles of NH₃ produced and distilled will be

$$\begin{gathered} =\left(0.214-0.0594\right)mmol \\ =0.1546mmol \end{gathered}$$

As 1 mol of Nitrogen (N) in the protein generates 1 mol of NH₃, there must have been 0.1546 mmol of Nitrogen in the protein, which corresponds to

$$\begin{gathered} =0.1546mmol×14.007\frac{mgN}{mmol} \\ =2.165mg \end{gathered}$$

As it is given that the protein contains 16.2 wt% of nitrogen, then the amount of protein will be

$$\begin{gathered} =\frac{2.165mgN}{0.162mgN/mgprotein} \\ =13.36mgprotein \end{gathered}$$

The amount of protein is present in 0.5 mL of solution

Therefore, actual concentration of protein will be

$$\begin{gathered} =\frac{13.36mgprotein}{0.500mL} \\ =26.72\frac{mgprotein}{mL} \end{gathered}$$