During determination of N₂ . It is seen that involved NH₃ from 3.5gm anniline can be neutralize by 3.69 cm³ of 0.0103 M HCl. The percentage of the N₂ in aniline is-

দেয়া আছে,

$\begin{array}{l} \text {অ্যানিলিন ভর }\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)=3.5 \mathrm{~g} \\ \mathrm{HCl} \text { আয়তন}=3.69 \mathrm{~cm}^{3} \\ \mathrm{HCl} \text { ঘনমাত্রা }=0.0103 \mathrm{~m} \\ n_{\text {HCl }}=V \times S=369 \times 10^{3} \times 0.0103 \\ =0.0381 \mathrm{mmol} \\ W=n \times m_{A}=n \times 14 \text { [ } N_{2} \text { আণবিক ভর ] } \\ =5.32 \times 10^{-4} \mathrm{~g} \text { [n মোল ] } \\ \therefore N_{2} \text { এর শতকরা পরিমাণ হলো } \text { }=\frac{5.32 \times 10^{-4}}{3.51} \times 100 \\ =0.01516 \% \\ \end{array}$